On the other hand, Eq. Rotational motion exists everywhere in the universe. Fig. What do you understand by the angular velocity of the wheel? The hinged door is a typical example. One example is rotation of an object flying freely in space which can rotate about the center of mass with any orientation. View ROTATION ABOUT A FIXED AXIS.pptx from EE 20224 at University of Notre Dame. The parameters that govern the rotational motion of a rigid body are angular displacement, angular velocity, and angular acceleration. The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. 0000004937 00000 n If the angular speed of the cylinder is 5 \(\mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}{:} (\mathrm {a})\) calculate the angular momentum of the cylinder about its central axis; (b) Suppose the cylinder accelerates at a constant rate of 0.5 \(\mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2}\), find the angular momentum of the cylinder at \(t=3\mathrm {s}(\mathrm {c})\) find the applied torque; (d) find the work done after \(3\mathrm {s}.\), (a) The moment of inertia of the cylinder is, for homogeneous symmetrical objects the total angular momentum is. Ropes wrapped around the inner and outer sections exert different forces, A block of mass m is attached to a light string that is wrapped around the rim of a uniform solid disk of radius R and mass M. Find the net torque on the system shown in Fig. If its angular acceleration is given by \(\alpha =(4t)\,\mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2}\) and if at \(t=0, \omega _{0}=0\), find the angular momentum of the sphere and the applied torque as a function of time. Suppose that the cylinder is free to rotate about its central axis and that the rope is pulled from rest with a constant force of magnitude of 35 N. Assuming that the rope does not slip, find: (a) the torque applied to the cylinder about its central axis; (b) the angular acceleration of the cylinder; (c) the acceleration of a point in the unwinding rope; (d) the number of revolutions made by the cylinder when it reaches an angular velocity of 12 \(\mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}, (\mathrm {e})\) the work done by the applied force when the rope is pulled a distance of \(1\mathrm {m}, (\mathrm {f})\) the work done using the workenergy theorem. 0000006896 00000 n 7.1 shows the linear/rotational analogous equations. It is named after Thomas Young. You'll recall from freshman physics that the angular momentum and rotational energy are L z = I , E r o t = 1 2 I 2 where (24.3.1) I = i m i r i 2 = d x d y d z ( x, y, z) r 2 Find: (a) the rotational kinetic energy of the disc; (b) Suppose that the same disc rotate using a motor that delivers an instantaneous of power 0. 15.1C Equations Defining the Rotation of a Rigid Body About a Fixed Axis Motion of a rigid body rotating around a fixed axis is often specified by the type of angular acceleration. And there will be the instantaneous angular velocity vector which is neither space- nor body-fixed. A body of mass m moving with velocity v has a kinetic energy of mv 2. Suppose a rigid body of an arbitrary shape is in pure rotational motion about the \(\mathrm {z}\)-axis (see Fig. A 5 kg uniform solid cylinder of radius 0.2 \(\mathrm {m}\) rotate about its center of mass axis with an angular speed of 10 rev/min. 0000005734 00000 n Hence, the total torque acting on the cylinder is, (b) The moment of inertia of the cylinder is. Therefore, by using the definition of vector product we may write, From Sect. This a detailed diagram of the same scenario. 1. A disc of radius 2.2 \(\mathrm {m}\) and mass of 120 kg rotate about a frictionless vertical axle that passes through its center. \label{jpb:eqn:5} The pulley comes to rest (momentarily) when $\omega=0$. Related . Let $I$ be the moment of inertia about the axis of rotation. Any point of the rotating body has a (linear) velocity, which at every moment of time is exactly the same as if the body were rotating around an axis directed along the angular velocity vector. The axis referred to here is the rotation axis of the tensor . \(\displaystyle \triangle L=\int _{t_{1}}^{t_{2}}\tau dt=\tau _{ave}\triangle t=\overline{F}Rt=(100 \; \mathrm {N})(0.2 \; \mathrm {m})(2\times 10^{-3} \; \mathrm {s})=0.04 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2}/\mathrm {s}\), That gives \(\omega _{f}=5.2 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}.\). In: Principles of Mechanics. We treat the whole system as a single point-like particle of mass m located at the center . 0000002657 00000 n axis of rotation : the straight line through all fixed points of a rotating rigid body around which all other points of the body move in circles Love words? The quantities \(\theta , \omega \) and \(\alpha \) in pure rotational motion are the rotational analog of x,v and a in translational one-dimensional motion. Problems involving the kinetics of a rigid body rotating about a fixed axis can be solved using the following process. Note that the concept of perfect rigidity has limitations in the theory of relativity since information cannot travel faster than the velocity of light, and thus signals cannot be transmitted instantaneously between the ends of a rigid body which is implied if the body had perfect rigidity. To simplify these problems, we define the translational and rotational motion of the body separately. If the particle undergoes this angular displacement during a time interval \(\triangle t\), the average angular velocity \(\overline{\omega }\) is then definedas, A rigid body of an arbitrary shape is in pure rotational motion about the \(\mathrm {z}\)-axis, The motion of a particle that lies in a slice of the body in the x-y plane, The particle is at point \(P_{1}\) at \(t_{1}\) and at \(P_{2}\) at \(t_{2}\), where it changes its angular position from \(\theta _{1}\) to \(\theta _{2}\), \(\omega \) has units of \(\mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}\) or \(\mathrm {s}^{-1}\). Principles of Mechanics pp 103122Cite as, Part of the Advances in Science, Technology & Innovation book series (ASTI). Substitute $\vec{a}$ from the previous equation into the last equation to get $F_x=-F/4$ and $F_y=\sqrt{3}m\omega^2l$. A uniform solid sphere of radius of 0.2 \(\mathrm {m}\) is rotating about its central axis with an angular speed of 5 \(\mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}\). For a rigid body which is a continuous system of particles, the sum is replaced by an integral. This decrease in kinetic energy is due to the internal nonconservative (frictional) force that acts within the system. When another disc of moment of inertia of 0.05 kg m\(^{2}\) that is initially at rest is dropped on the first, the two will eventually rotate with the same angular speed due to friction between them. \begin{align} Going by this logic, raw egg should stop first if frictional forces are equal in two cases. In solving problems \(\rho , \sigma \), and \(\lambda \) (see Sect. For a rigid body undergoing fixed axis rotation about the center of mass, our rotational equation of motion is similar to one we have already encountered for fixed axis rotation, ext = dLspin / dt . In contrast, when the torque acting on a body produces angular acceleration, it is called dynamic torque. 0000003918 00000 n In other words, the speed depends on the torque applied to the door. If the rod is released from rest at an angle \(\theta =30^{\mathrm {o}}\) to the horizontal, find; (a) the initial angular acceleration of the rod when it is released; (b) the initial acceleration of a point at the end of the rod; (c) from conservation of energy find the angular speed of the rod at its lowest position (Neglect friction at the pivot). where I is the moment of inertia of the rigid body about the rotational axis (z-axis). 8.4). But we must first understand rotational motion and its nuances. As seen from Fig. Since A is a fixed point, the torque $\tau$ is related to the angular acceleration $\alpha$ by The hinged door is a typical example. This is followed by a discussion of practical applications. Angular Displacement The vectors \(\omega \) and \(\alpha \) are not used in the case of pure rotational motion, they are used in the general rotational motion when the axis of rotation changes its direction with time. The angular displacement of the particle is related to s by, where r is the radius of the circle in which the particle is moving along. It is shown that the angular momentum (torque) and angular velocity (acceleration) vectors are parallel to each other if the fixed reference point is chosen as follows: (i) for a body of arbitrary shape rotating about a . Pretend that you are an observer at . Young's modulus is a measure of the elasticity or extension of a material when it's in the form of a stressstrain diagram. A 5 kg wheel of radius of 0.1 \(\mathrm {m}\) decelerates from an angular speed of 5 \(\mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}\) to rest after going through an angular displacement of 10 rev If a frictional force causes the wheel to decelerate, find the torque due to this force. where $I=10$ kg-m2 is moment of inertia of the pulley. The discussion of general rotation, in which both the position and the direction of the axis change, is quite complex. (c) After seven revolutions the angular velocity is, that gives \(\omega =16.24 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}\). Pages 1 Ratings 100% (1) 1 out of 1 people found this document helpful; This . 7.32). 7.6). \end{align} \end{align} Anyone you share the following link with will be able to read this content: Sorry, a shareable link is not currently available for this article. Accessibility StatementFor more information contact us atinfo@libretexts.orgor check out our status page at https://status.libretexts.org. At any given point, the tangent to a specific point denotes the angular velocity of a body. Because the origin is taken at the center of mass we have, The moment of inertia of the object about the center of mass axis is, where x and y are the coordinates of the mass element dm from the center of mass (the origin). 7.10. A body in rotational motion can be rotating around a fixed axis or a fixed point. \end{align} 7.18, then each volume element is given by, Method 2: Using double integration: dividing the cylinder into thin rods each of mass, Method 3: Using triple integration Dividing the cylinder into small cubes each of mass given by. \vec{a}&=a_x\,\hat\imath+a_y\,\hat\jmath \\ Thus, to find the rotational inertia, the axis of rotation must be specified. Objects are made up of particles. Therefore the net external torque is, The moment of inertia about the rotational axis is, (b) The acceleration of a point at the end of the rod is, (c) When the rod reaches its lowest position, the potential energy of its center of mass is transformed into rotational kinetic energy of the rod. The angular position of P is defined by . Angular velocity, , is . Chapter 12 Rotation of a Rigid Body. Fixed-axis rotation describes the rotation around a fixed axis of a rigid body; that is, an object that does not . A system of particles rotating about the z-axis, The parallel-axis theorem states that the moment of inertia I of a system about any axis that is parallel to an axis passing through the center of mass is. If a force that lies in the x-y plane is applied to the body at \(\mathrm {P}\), then the work done on the body if it rotates through an angle \( d\theta \) is, Since \(\varvec{\tau }\) and \(\varvec{\omega }\) are parallel, (the force lies in the x-y plane therefore the total torque is parallel to the \(\mathrm {z}\)-axis) we have, Therefore, the total work done in displacing the body from \(\theta _{1}\) to \(\theta _{2}\) is, The WorkEnergy Theorem The workenergy theorem states that the work done by an external force while a rigid object rotate from \(\theta _{1}\) to \(\theta _{2}\) is equal to the change in the rotational energy of the object. Force is responsible for all motion that we observe in the physical world. Fig. When force is applied, the door rotates. The simplest case is pictured above, a single tiny mass moving with a constant linear velocity (in a straight line.) 7.16 shows a uniform thin rod of mass M and length L. Find the moment of inertia of the rod about an axis that is perpendicular to it and passing through: (a) the center of mass; (b) at one end; (c) at a distance of L/6 from one end. 7.29. The general plane motion: The motion here can be considered as a combination of pure translational motion parallel to a fixed plane in addition to a pure rotational motion about an axis that is perpendicular to that plane. If \(m=0.1 \; \mathrm {k}\mathrm {g},\) find the moment of inertia of the system and the corresponding kinetic energy if it rotates with an angular speed of 5 \(\mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}\) about: (a) the \(\mathrm {z}\)-axis; (b) the \(\mathrm {y}\)-axis and; (c) the \(\mathrm {x}\)-axis \((a=0.2 \; \mathrm {m})\). Thus the best solution for describing rotation of a rigid body is to use a rotation matrix that transforms from the stationary fixed frame to the instantaneous body-fixed frame for which the moment of inertia tensor can be evaluated. Search: Opencv Rigid Transform. If a rigid body is rotating about a fixed axis, the particles will follow a circular path. In spite of this, the pencil always has the same unique inertia tensor in the body-fixed frame. 1 APPLICATIONS The crank on the oil-pump rig undergoes rotation about a fixed axis, caused by the driving torque M from a motor. Dynamics Of Rotational Motion About A Fixed Axis Rigid bodies undergo translational as well as rotational motion. $$ 1 \; \text {rev} =360^{\circ }=2\pi \; \text {rad} $$, $$ 1 \; \text {rad} =57.3^{\circ }=0.159 \; \text {rev} $$, \(2\pi \mathrm {r}\mathrm {a}\mathrm {d}\), $$ \theta =(2\pi +2\pi +2\pi ) \; \text {rad} =6\pi \; \text {rad} $$, $$ \triangle \theta =\theta _{2}-\theta _{1} $$, $$ \overline{\omega }=\frac{\theta _{2}-\theta _{1}}{t_{2}-t_{1}}=\frac{\triangle \theta }{\triangle t} $$, $$ \omega =\lim _{\triangle t\rightarrow 0}\frac{\triangle \theta }{\triangle t}=\frac{d\theta }{dt} $$, \(\mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}\), $$ \overline{\alpha }=\frac{\omega _{2}-\omega _{1}}{t_{2}-t_{1}}=\frac{\triangle \omega }{\triangle t} $$, $$ \alpha =\lim _{\triangle t\rightarrow 0}\frac{\triangle \omega }{\triangle t}=\frac{d\omega }{dt} $$, \(\mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2}\), \(\mathrm {r}\mathrm {e}\mathrm {v}/\mathrm {s}^{2}\), \(\mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}.\), $$ 15^{\mathrm {o}}=(15 \; \displaystyle \mathrm {deg})\bigg (\frac{1 \; \mathrm {r}\mathrm {e}\mathrm {v}}{360 \; \mathrm {deg}}\bigg )=0.042 \; \text {rev} $$, $$ 15^{\mathrm {o}}=(15 \; \displaystyle \mathrm {deg})\bigg (\frac{2 \; \pi \mathrm {r}\mathrm {a}\mathrm {d}}{360 \; \mathrm {deg}}\bigg )=0.26 \; \text {rad} $$, $$ 0.25 \; \displaystyle \mathrm {r}\mathrm {e}\mathrm {v}/\mathrm {s}^{2}=\bigg (0.25 \; \frac{\mathrm {r}\mathrm {e}\mathrm {v}}{\mathrm {s}^{2}}\bigg )\bigg (\frac{2\pi \; \mathrm {r}\mathrm {a}\mathrm {d}}{1\mathrm {r}\mathrm {e}\mathrm {v}}\bigg )=1.57 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2} $$, $$ 0.25 \; \displaystyle \mathrm {r}\mathrm {e}\mathrm {v}/\mathrm {s}^{2}=\bigg (0.25 \; \frac{\mathrm {r}\mathrm {e}\mathrm {v}}{\mathrm {s}^{2}}\bigg )\bigg (\frac{360 \; \mathrm {deg}}{1\,\mathrm {r}\mathrm {e}\mathrm {v}}\bigg )=90 \; \mathrm {deg}/\mathrm {s}^{2} $$, $$ 3\ \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}=\bigg (3 \; \frac{\mathrm {r}\mathrm {a}\mathrm {d}}{\mathrm {s}}\bigg )\bigg (\frac{1 \; \mathrm {r}\mathrm {e}\mathrm {v}}{2\pi \; \mathrm {r}\mathrm {a}\mathrm {d}}\bigg )=0.48 \; \mathrm {r}\mathrm {e}\mathrm {v}/\mathrm {s} $$, $$ 3 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}=\bigg (3 \; \frac{\mathrm {r}\mathrm {a}\mathrm {d}}{\mathrm {s}}\bigg )\bigg (\frac{360^{\mathrm {o}} \; \mathrm {deg}}{2\pi \; \mathrm {r}\mathrm {a}\mathrm {d}}\bigg )=172 \; \mathrm {deg}/\mathrm {s} $$, $$ \theta _{1}=((0.3)(1 \; \mathrm {s})^{2}+(0.4)(1 \; \mathrm {s})^{3})=0.7 \; \text {rad} $$, $$ \theta _{2}=((0.3)(2 \; \mathrm {s})^{2}+(0.4)(2 \; \mathrm {s})^{3})=4.4 \; \text {rad} $$, $$ \triangle \theta =( 4.4 \; \text {rad})-(0.7\,\text {rad}) =3.7 \; \text {rad} $$, $$ \overline{\omega }=\frac{\triangle \theta }{\triangle t}=\frac{(3.7 \; \mathrm {r}\mathrm {a}\mathrm {d})}{(1 \; \mathrm {s})}=3.7 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s} $$, $$ \omega =\frac{d\theta }{dt}=((0.6)t+(1.2)t^{2}) \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s} $$, $$ \omega =(0.6)(5 \; \mathrm {s})+(1.2)(5 \; \mathrm {s})^{2}=33 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s} $$, $$ \alpha =\frac{d\omega }{dt}=((0.6)+(2.4)t) \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2} $$, $$ \alpha =(0.6)+(2.4)(5 \; \mathrm {s})=12.6 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2} $$, \(\alpha =(9-2t) \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2}\), $$ \omega =\int \alpha dt=\int (9-2t)dt=9t-t^{2}+c_{1} $$, \(\omega =2 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}\), \(c_{1}=2 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}\), $$ \omega =(9t-t^{2}+2) \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s} $$, $$ \theta =\int \omega dt=\int (9t-t^{2}+2)dt=\frac{9}{2}t^{2}-\frac{1}{3}t^{3}+2t+c_{2} $$, \(t=0, \theta =3 \; \mathrm {r}\mathrm {a}\mathrm {d}\), $$ \displaystyle \theta =\bigg (\frac{9}{2}t^{2}-\frac{1}{3}t^{3}+2t+3 \bigg ) \; \text {rad} $$, $$ \displaystyle \theta =\frac{9}{2}(4.5 \; \mathrm {s})^{2}-\frac{1}{3}(4.5 \; \mathrm {s})^{3}+2(4.5 \; \mathrm {s})+3=72.8 \; \text {rad} $$, \(t_{1}=0, t_{2}=t, \omega _{1}=\omega _{\mathrm {o}}, \omega _{2}=\omega , \theta _{1}=\theta _{\mathrm {o}}\), $$ \overline{\omega }=\frac{\omega _{0}+\omega }{2} $$, $$ \alpha =\overline{\alpha }=\frac{\omega _{2}-\omega _{1}}{t_{2}-t_{1}}=\frac{\omega -\omega _{0}}{t} $$, $$\begin{aligned} \omega =\omega _{0}+\alpha t \end{aligned}$$, $$ \overline{\omega }=\frac{\theta _{2}-\theta _{1}}{t_{2}-t_{1}}=\frac{\theta -\theta _{0}}{t}=\frac{\omega _{0}+\omega }{2} $$, $$\begin{aligned} \displaystyle \theta =\theta _{0}+\frac{1}{2}(\omega _{0}+\omega )t \end{aligned}$$, $$ \theta =\theta _{0}+\frac{1}{2}(\omega _{0}+\omega )t=\theta _{0}+\frac{1}{2}(\omega _{0}+\omega _{0}+\alpha t)t $$, $$\begin{aligned} \displaystyle \theta =\theta _{0}+\omega _{0}t+\frac{1}{2}\alpha t^{2} \end{aligned}$$, $$ \theta =\theta _{0}+\frac{1}{2}(\omega _{0}+\omega )t=\theta _{0}+\frac{1}{2}(\omega _{0}+\omega )\left( \frac{\omega -\omega _{0}}{\alpha }\right) $$, $$\begin{aligned} \omega ^{2}=\omega _{0}^{2}+2\alpha (\theta -\theta _{0}) \end{aligned}$$, $$\begin{aligned} v=r\omega \end{aligned}$$, $$\begin{aligned} a_{r}=\displaystyle \frac{v^{2}}{r} \end{aligned}$$, $$ a_{t}=\frac{dv}{dt}=r\frac{d\omega }{dt} $$, $$ \mathbf {a}=\mathbf {a}_t+\mathbf {a}_r $$, $$ a=\sqrt{{a_t}^2+{a_r}^2}=\sqrt{{r}^2{\alpha }^2+{r}^2{\omega }^4}=r\sqrt{{\alpha }^2+{\omega }^4} $$, \(\theta =(60\;\mathrm {deg})(2\pi \mathrm {r}\mathrm {a}\mathrm {d}/360\;\mathrm {deg})=1.05\), $$ \theta =\theta _{0}+\omega _{0}t+\frac{1}{2}\alpha t^{2} $$, $$ \alpha =\frac{2\theta }{t^{2}}=\frac{2(1.05 \; \mathrm {r}\mathrm {a}\mathrm {d})}{(2 \; \mathrm {s})^{2}}=0.525 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2} $$, $$ \omega =\omega _{0}+\alpha t=(0.525 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2})(2 \; \mathrm {s})=1.05 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s} $$, $$ \omega =(0.525 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2})(6\mathrm {s})=3.15 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s} $$, $$ v=r\omega =(0.07 \; \mathrm {m})(1.05 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})=0.074 \; \mathrm {m}/\mathrm {s} $$, $$ s=r\theta =(0.07 \; \mathrm {m})(1.05 \; \mathrm {rad} ) =0.074 \; \mathrm {m} $$, \(\mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s},\), $$ r_{1}\omega _{1}=r_{2}\omega _{2}=v $$, $$ \omega _{2}=\frac{r_{1}}{r_{2}}\omega _{1}=\frac{(2 \; \mathrm {c}\mathrm {m})}{(5 \; \mathrm {c}\mathrm {m})}(2 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})=0.8 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s} $$, $$ \displaystyle \omega =rv=\frac{2\pi }{T}=\frac{2(3.14)}{(27.3 \; \mathrm {d}\mathrm {a}\mathrm {y})}=0.23 \; \text {rad/day} $$, $$ \displaystyle \omega = \bigg (0.23 \; \frac{\mathrm {r}\mathrm {a}\mathrm {d}}{\mathrm {d}\mathrm {a}\mathrm {y}}\bigg ) \bigg (\frac{1 \; \mathrm {r}\mathrm {e}\mathrm {v}}{2\pi \; \mathrm {r}\mathrm {a}\mathrm {d}}\bigg )=0.037 \; \text {rev/day} $$, $$\begin{aligned} \mathbf {v}=\varvec{\omega }\times \mathbf {R} \end{aligned}$$, $$ \mathbf {a}=\frac{d\mathbf {v}}{dt}=\frac{d}{dt}(\varvec{\omega }\times \mathbf {R}) $$, \((d/dt(\mathbf {A}\times \mathbf {B})=\mathbf {A}\times d\mathbf {B}/dt+d\mathbf {A}/dt\times \mathbf {B})\), $$ \mathbf {a}=\frac{d \varvec{\omega }}{dt}\times \mathbf {R}+\varvec{\omega }\times \frac{d\mathbf {R}}{dt} $$, $$ =\varvec{\alpha }\times \mathbf {R}+\varvec{\omega }\times \mathbf {v} $$, $$ |\varvec{\alpha }\times \mathbf {R}|=\alpha R\sin \theta =r\alpha =a_{t} $$, $$\begin{aligned} \mathbf {a_{t}}=\varvec{\alpha }\times \mathbf {R} \end{aligned}$$, $$ |\varvec{\omega }\times \mathbf {v}|=\omega v\sin 90^{\mathrm {o}}=\omega v=r\omega ^{2}=a_{r} $$, $$\begin{aligned} \mathbf {a}_{r}=\varvec{\omega }\times \mathbf {v} \end{aligned}$$, \(K=\displaystyle \frac{1}{2}\sum _{i}m_{i}v_{i}^{2}\), $$ K_{R}=\frac{1}{2}\sum _{i}(m_{i}r_{i}^{2})\omega ^{2} $$, $$ I=\lim _{\triangle m_{\mathrm {i}\rightarrow 0}}\sum _{i}m_{i}r_{i}^{2}=\int r^{2}dm $$, $$ I_{cm}=\int r^{2}dm=\int (x^{2}+y^{2})dm $$, $$ I_{P}=\int [(x-x_{P})^{2}+(y-y_{P})^{2}]dm $$, $$ I_{P}=\int (x^{2}+y^{2})dm-2x_{P}\int xdm-2y_{P}\int ydm+\int (x_{P}^{2}+y_{P}^{2})dm $$, $$ I_{P}=I_{cm}+MD^{2} \quad \text {(Parallel--Axis Theorem)} $$, $$ \mathbf {L}_{i}=\mathbf {R}_{i}\times \mathbf {p}_{i} $$, $$ L_{iz}=L_{i}\sin \theta =R_{i}p_{i}\sin \theta =R_{i} ({ m_{i} v_{i}}) \sin \theta $$, $$ =R_{i}m_{i}(r_{i}\omega )\sin \theta =m_{i}r_{i}^{2}\omega $$, $$ L_{z}=\sum _{i}m_{i}r_{i}^{2}\omega =\bigg (\sum _{i}m_{i}r_{i}^{2}\bigg )\omega $$, $$\begin{aligned} \mathbf {L}_{z}=I\varvec{\omega } \end{aligned}$$, $$ \mathbf {L}=\sum _{i}\mathbf {L}_{iz}=\mathbf {L}_{z}=I\varvec{\omega }$$, $$\begin{aligned} \mathbf {L}=I\varvec{\omega } \end{aligned}$$, $$ \Sigma {\varvec{\tau }_{ext}}=\frac{d\mathbf {L}}{dt} $$, $$ \Sigma {\varvec{\tau }_{extz}}=\frac{d\mathbf {L}_{z}}{dt}=\frac{d(I\varvec{\omega })}{dt}=I\varvec{\alpha }$$, $$ \Sigma {\varvec{\tau }_{ext}}=\frac{d\mathbf {L}}{dt}=\frac{d(I\varvec{\omega })}{dt}=I\varvec{\alpha }$$, $$ \triangle \theta =( 10 \; \mathrm {rev}) \bigg (\displaystyle \frac{2\pi \; \mathrm {r}\mathrm {a}\mathrm {d}}{1 \; \mathrm {r}\mathrm {e}\mathrm {v}}\bigg )=62.8 \; \text {rad} $$, $$ \alpha =\frac{\omega ^{2}-\omega _{0}^{2}}{2\triangle \theta }=\frac{0-(5 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})^{2}}{2(62.8 \; \mathrm {r}\mathrm {a}\mathrm {d})}=-0.2 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s} $$, $$ \tau =I\alpha =MR^{2}\alpha =(5 \; \mathrm {k}\mathrm {g})(0.1 \; \mathrm {m})^{2}(-0.2 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2})=-0.01 \; \mathrm {N}\,\mathrm {m} $$, $$\begin{aligned} I_{z}&=\displaystyle \sum _{i}m_{i}r_{i}^{2}=2ma^{2}+\frac{m}{2}a^{2}+ma^{2}=\frac{7}{2}ma^{2}\\&=\frac{7}{2}(0.1 \; \mathrm {k}\mathrm {g})(0.2 \; \mathrm {m})^{2}=0.014 \; \mathrm {kg\, m^2} \end{aligned}$$, $$ K_{R}=\frac{1}{2}I_{z}\omega ^{2}=\frac{1}{2}(0.014 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})(5 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})^{2}=0.175 \; \mathrm {J} $$, $$I_{y}=\displaystyle \frac{m}{2}a^{2}+2ma^{2}=\frac{5}{2}ma^{2}=\frac{5}{2}(0.1 \; \mathrm {k}\mathrm {g})(0.2 \; \mathrm {m})^{2}=0.01 \; \mathrm {kg\, m^2}$$, $$ K_{R}=\frac{1}{2}I_{y}\omega ^{2}=\frac{1}{2}(0.01 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})(5 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})^{2}=0.125 \; \mathrm {J} $$, $$ I_{x}=ma^{2}=(0.1 \; \mathrm {k}\mathrm {g})(0.2 \; \mathrm {m})^{2}=4\times 10^{-3} \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2} $$, $$ K_{R}=\frac{1}{2}I_{x}\omega ^{2}=\frac{1}{2}(4\times 10^{-3} \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})(5 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})^{2}=0.05 \; \mathrm {J} $$, $$ dm=\lambda dx=\bigg (\frac{M}{L}\bigg )dx $$, $$ I_{cm}=I_{y}=\displaystyle \int r^{2}dm=\int _{x=-\frac{L}{2}}^{\frac{L}{2}}x^{2}\bigg (\frac{M}{L}\bigg )dx=\frac{M}{L}\bigg (\frac{x^{3}}{3}\bigg ) \bigg |_{-L/2}^{L/2}=\displaystyle \frac{1}{12}ML^{2} $$, $$ I_{y'}=I_{cm}+MD^{2}=\frac{1}{12}ML^{2}+M\bigg (\frac{L}{2}\bigg )^{2}=\frac{1}{3}ML^{2} $$, $$ I_{y''}=I_{cm}+MD^{2}=\frac{1}{12}ML^{2}+M\bigg (\frac{L}{2}-\frac{L}{6}\bigg )^{2}=\frac{7}{36}ML^{2} $$, $$ I_{cm}=\int r^{2}dm=\int r^{2}\sigma dA=\int _{y=-a/2}^{a/2}\int _{y=-b/2}^{b/2}{(x^{2}+y^{2})\bigg (\frac{M}{ab}\bigg )dxdy} $$, $$ =\displaystyle \frac{M}{ab} \int _{y=-a/2}^{a/2} {{\bigg (\frac{x^3}{3}+xy^2\bigg )|_{x=-b/2}^{b/2}}dy}=\frac{M}{ab} \int _{y=-a/2}^{a/2} {{\bigg (\frac{b^3}{12}+by^2\bigg )}dy} $$, $$ =\displaystyle \frac{M}{ab} \bigg (\frac{b^3y}{12}+\frac{y^3b}{3}\bigg ) \bigg |_{x=-a/2}^{a/2}=\frac{M}{ab} \bigg [\frac{ab^3}{12}+\frac{ab^3}{12}\bigg ]=\frac{1}{12}M\big (a^2+b^2\bigg ) $$, $$ I=\int r^{2}dm=\int _{0}^{R}r^{2}(\rho 2\pi rLdr)=2\pi \rho L\int _{0}^{R}r^{3}dr=\frac{\pi \rho L}{2}R^{4} $$, $$ I=\int r^{2}dm=\int _{0}^{2\pi } \int _{r=0}^{R} r^{3}\rho Ldrd\theta =\rho \frac{L}{4}R^{4}\int _{\theta =0}^{2\pi } d\theta =\frac{\pi \rho LR^{4}}{2} $$, $$ I=\int r^{2}dm= \int _{\theta =0}^{2\pi } \int _{r=0}^{R} \int _{z=0}^{L}\rho r^{3}drd\theta dz=\rho L\frac{R^{4}}{4}\int _{\theta =0}^{2\pi } d\theta =\frac{\pi \rho LR^{4}}{2} $$, $$ I=I_{1}+I_{2}+I_{3}=3\bigg (\frac{1}{3}ML^{2}\bigg )=ML^{2} $$, $$ dA=2\pi R\sin \theta Rd\theta =2\pi R^{2}\sin \theta d\theta $$, $$ I=\int r^{2}dm=\int R^{2}\sin ^{2}\theta \sigma 2\pi R^{2}\sin \theta d\theta $$, $$ I=\frac{M}{2}R^{2}\int _{\theta =0}^{\pi }\sin ^{3}\theta d\theta =\frac{M}{2}R^{2}\int _{\theta =0}^{\pi }(1-\cos ^{2}\theta )\sin \theta d\theta $$, $$ =\frac{M}{2}R^{2}\bigg [-\cos \theta +\frac{\cos ^{3}\theta }{3}\bigg ]_{\theta =0}^{\pi }=\frac{2}{3}MR^{2} $$, $$ \mathbf {L}_{i}=\mathbf {L}_{f}= \mathrm{constant~(isolated~system)} $$, $$ dW=\mathbf {F}\cdot d\mathbf {s}=\mathbf {F}\cdot \frac{d\mathbf {s}}{dt}dt=\mathbf {F}\cdot \mathbf {v} dt=\mathbf {F}\cdot (\varvec{\omega }\times \mathbf {r})dt $$, $$ =(\mathbf {r}\times \mathbf {F})\cdot \varvec{\omega }dt=\varvec{\tau }\cdot \varvec{\omega }dt $$, $$ dW=\tau \omega dt=\tau \frac{d\theta }{dt}dt=\tau d\theta $$, $$\begin{aligned} W=\displaystyle \int _{\theta _{1}}^{\theta _{2}}\tau d\theta \end{aligned}$$, $$ W=\tau (\theta _{2}-\theta _{1})=\tau \triangle \theta $$, $$ W=\int _{\theta _{1}}^{\theta _{2}}\tau d\theta =\int _{\theta _{1}}^{\theta _{2}}I\alpha d\theta =\int _{\omega _{1}}^{\omega _{2}}I\omega \frac{d\omega }{dt}dt=\int _{\omega _{1}}^{\omega _{2}}I\omega d\omega =\frac{1}{2}I\omega _{2}^{2}-\frac{1}{2}I\omega _{1}^{2} $$, $$ W=\triangle K=\frac{1}{2}I\omega _{2}^{2}-\frac{1}{2}I\omega _{1}^{2} $$, $$ P=\frac{dW}{dt}=\frac{\tau _{z}d\theta }{dt}=\tau _{z}\omega _{z} $$, $$ I=\displaystyle \frac{1}{2}MR^{2}=\frac{1}{2}(5 \; \mathrm {k}\mathrm {g})(0.08 \; \mathrm {m})^{2}=0.016 \; \mathrm {kg\, m^2} $$, $$ \omega =\bigg (\frac{170 \; \mathrm {r}\mathrm {e}\mathrm {v}}{\min }\bigg )\bigg (\frac{2\pi \; \mathrm {r}\mathrm {a}\mathrm {d}}{1 \; \mathrm {r}\mathrm {e}\mathrm {v}}\bigg )\bigg (\frac{1 \; \min }{60 \; \mathrm {s}}\bigg )=17.8 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s} $$, $$ K=\frac{1}{2}I\omega ^{2}=\frac{1}{2}(0.016 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})(17.8 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})^{2}=2.5 \; \mathrm {J} $$, $$ P=(0.2 \; \mathrm {h}\mathrm {p}\ ) \bigg (\frac{746 \; \mathrm {W}}{1\mathrm {h}\mathrm {p}}\bigg )=149.2 \; \mathrm {W} $$, $$ \tau =\frac{P}{\omega }=\frac{(149.2 \; \mathrm {W})}{(17.8 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})}=8.4 \; \mathrm {N}\,\mathrm {m} $$, \(\mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}, (\mathrm {e})\), $$ \tau =FR=(35 \; \mathrm {N})(0.2 \; \mathrm {m})=7 \; \text {N/m} $$, $$ I=MR^{2}=(30 \; \mathrm {k}\mathrm {g})(0.2 \; \mathrm {m})^{2}=1.2 \; \mathrm {kg\, m^2} $$, $$ \alpha =\frac{\tau }{I}=\frac{(7 \; \mathrm {N}\,\mathrm {m})}{(1.2\;\mathrm {k}\mathrm {g}\,\mathrm {m}^{2})}=5.8 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2} $$, $$ a=R\alpha =(0.2 \; \mathrm {m})(5.8 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2})=1.2 \; \mathrm {m}/\mathrm {s}^{2} $$, $$ \omega ^{2}=\omega _{0}^{2}+2\alpha \theta $$, $$ \displaystyle \theta =\frac{(12\,\mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})^{2}}{2(5.8\,\mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2})}=12.4 \; \mathrm {rad} $$, $$ \theta =(12.4 \; \mathrm {rad}) \bigg (\displaystyle \frac{1 \; \mathrm {r}\mathrm {e}\mathrm {v}}{2\pi \; \mathrm {r}\mathrm {a}\mathrm {d}}\bigg )= 2 \; \text {rev} $$, $$ \displaystyle \theta =\frac{s}{R}=\frac{(1 \; \mathrm {m})}{(0.2 \; \mathrm {m})}=5 \; \mathrm {rad} $$, $$ W=\int _{\theta _{0}}^{\theta }\tau d\theta =\tau (\theta -\theta _{0})=(7 \; \mathrm {N}\,\mathrm {m})\ ((5 \; \mathrm {r}\mathrm {a}\mathrm {d})-0)=35 \; \mathrm {J} $$, $$ \omega ^{2}=\omega _{0}^{2}+2\alpha \theta =0+2(5.8 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2}) ( 5 \; \mathrm {rad}) $$, \(\omega =7.6 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}\), $$ W=\triangle K=\frac{1}{2}I\omega ^{2}-\frac{1}{2}I\omega _{0}^{2}=\frac{1}{2}(1.2 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})(7.6 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})^{2}-0=35 \; \mathrm {J} $$, $$\displaystyle \tau =\frac{MgL}{2}\cos \theta =\frac{(0.75\; \mathrm {k}\mathrm {g})(9.8 \; \mathrm {m}/\mathrm {s}^{2})(1 \; \mathrm {m})}{2}\cos 30^{\circ }=3.2\,\mathrm {N\, m}$$, $$ I=\frac{1}{3}ML^{2}=\frac{(0.75\;\mathrm {k}\mathrm {g})(1 \; \mathrm {m})^{2}}{3}=0.25 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2} $$, $$ \alpha =\frac{\tau }{I}=\frac{(3.2 \; \mathrm {N}\,\mathrm {m})}{(0.25 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})}=12.8 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2} $$, $$ a_{t}=r\alpha =L\alpha =(1 \; \mathrm {m})(12.8 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2})=12.8 \; \mathrm {m}/\mathrm {s}^{2} $$, $$ 0+Mg\frac{L}{2}(\sin \theta +1)=\frac{1}{2}I\omega ^{2}+0 $$, $$\omega =\sqrt{Mg\frac{L}{I}(\sin \theta +1)}=\sqrt{\frac{(0.75 \; \mathrm {k}\mathrm {g})(9.8 \; \mathrm {m}/\mathrm {s}^{2})(1\mathrm {m})}{(0.25 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})}(\sin 30^{\circ }+1)}=6.64 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}$$, \(F_{1}=10 \; \mathrm {N}, F_{2}=20 \; \mathrm {N}\), $$\begin{aligned} \tau _{\mathrm {n}\mathrm {e}\mathrm {t}}&=\tau _{1}+\tau _{2}+\tau _{3}=(10 \; \mathrm {N})(0.05\;\mathrm {m})+(20 \; \mathrm {N})(0.05 \; \mathrm {m})\\&-(15 \; \mathrm {N})(0.15 \; \mathrm {m})=-0.75 \; \mathrm {N\, m} \end{aligned}$$, $$\begin{aligned} a=\frac{mg-T}{m} \end{aligned}$$, $$\begin{aligned} a=R\displaystyle \alpha =\frac{TR^{2}}{I} \end{aligned}$$, $$ T=\frac{g}{1/m+R^{2}/I}=\frac{g}{1/m+2R^{2}/MR^{2}} $$, $$ a=\frac{TR^{2}}{I}=\frac{2TR^{2}}{MR^{2}} $$, $$ \alpha =\frac{a}{R}=\frac{g}{R(1+M/{2m})} $$, $$\displaystyle \tau =I\alpha =\frac{2}{5}MR^{2}\alpha =\frac{2}{5}(4.7 \; \mathrm {k}\mathrm {g})(0.05 \; \mathrm {m})^{2}(3\,\mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2})=0.014 \; \mathrm {N\,} $$, $$ \theta =( 7 \; \mathrm {rev}) \bigg (\displaystyle \frac{2\pi \mathrm {r}\mathrm {a}\mathrm {d}}{1 \; \mathrm {r}\mathrm {e}\mathrm {v}}\bigg )=44 \; \text{ rad } $$, $$ W=\tau \triangle \theta = (0.014 \; \mathrm {N/m}) (44 \; \mathrm {rad}) =0.6 \; \mathrm {J} $$, $$ \omega ^{2}=\omega _{0}^{2}+2\alpha (\theta -\theta _{0}) $$, $$ \omega ^{2}=2\alpha \theta =2(3 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2}) ( 44 \; \text {rad}) $$, \(\omega =16.24 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}\), $$ W=\frac{1}{2}I\omega ^{2}-\frac{1}{2}I\omega _{0}^{2}=\frac{1}{2}(4.7\times 10^{-3} \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})(16.24 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2})^{2}-0=0.6 \; \mathrm {J} $$, $$ T_{1}-T_{2}+g(m_{2}-m_{1})=a(m_{1}+m_{2}) $$, $$ T_{2}-T_{1}=\frac{I\alpha }{R}=\frac{Ia}{R^{2}} $$, $$ a=\frac{g(m_{2}-m_{1})}{(m_{1}+m_{2}+I/R^{2)}} $$, $$ a=\frac{g(m_{2}-m_{1})}{(m_{1}+m_{2}+M/{2})} $$, $$ \alpha =\frac{g(m_{2}-m_{1})}{R(m_{1}+m_{2}+M/2)} $$, \(\mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}{:} (\mathrm {a})\), $$ I=\displaystyle \frac{1}{2}MR^{2}=\frac{1}{2}(10 \; \mathrm {k}\mathrm {g})(0.2 \; \mathrm {m})^{2}=0.2 \; \mathrm {kg\, m^2} $$, $$ L=I\omega =(0.2 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})(5 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})=1 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2}/\mathrm {s} $$, $$ \omega =\omega _{0}+\alpha t=(5 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})+(0.5 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2})(3 \; \mathrm {s})=6.5 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s} $$, $$ L=I\omega =(0.2 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})(6.5 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})=1.3 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2}/\mathrm {s} $$, $$\tau =I\alpha =(0.2 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})(0.5 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2})=0.1 \; \mathrm {N\, m}$$, $$ W=\frac{1}{2}I\omega ^{2}-\frac{1}{2}I\omega _{0}^{2}=\frac{1}{2}(0.2\,\mathrm {k}\mathrm {g}\,\mathrm {m}^{2})((6.5 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})^{2}-(5 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})^{2})=1.72 \; \mathrm {J} $$, \(\alpha =(4t)\,\mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2}\), $$ \omega =\int \alpha dt=\int 4tdt=2t^{2}+c $$, $$ \omega =(2t^{2}) \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s} $$, $$ I=\displaystyle \frac{2}{5}MR^{2}+MR^{2}=\frac{7}{5}MR^{2}=\frac{7}{5}(4.7 \; \mathrm {k}\mathrm {g})(0.05 \; \mathrm {m})^{2}=0.016 \; \mathrm {kg\, m^2} $$, $$ L=I\omega =(0.016 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})((2t^{2}) \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})=(0.03t^{2}) \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2}/\mathrm {s} $$, $$ \displaystyle \tau =\frac{dL}{dt}=(0.06t) \mathrm {N\, m} $$, $$ L=I_{z}\omega =(0.014 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})(5 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})=0.07 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2}/\mathrm {s} $$, $$ L=I_{y}\omega =(0.01 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})(5 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})=0.05 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2}/\mathrm {s} $$, $$ L=I_{x}\omega =(4\times 10^{-3} \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})(5 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})=0.02 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2}/\mathrm {s} $$, $$ \triangle L=I(\omega _{f}-\omega _{i}) $$, $$ (0.04 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2}/\mathrm {s})=(0.2 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})(\omega _{f}-(5 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})) $$, \(\omega _{f}=5.2 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}.\), \(\mathrm {r}\mathrm {e}\mathrm {v}/\mathrm {s}{:}\,(\mathrm {a})\), \(I_{f}=3 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})\), $$ \omega _{f}=\frac{I_{i}}{I_{f}}\omega _{i}=\frac{(15 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2}/\mathrm {s})}{(3 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2}/\mathrm {s})}(0.3 \; \mathrm {r}\mathrm {e}\mathrm {v}/\mathrm {s})=1.5 \; \mathrm {r}\mathrm {e}\mathrm {v}/\mathrm {s} $$, $$ \omega _{i}=\bigg (0.3 \; \frac{\mathrm {r}\mathrm {e}\mathrm {v}}{\mathrm {s}}\bigg ) \bigg (\frac{2\pi \; \mathrm {r}\mathrm {a}\mathrm {d}}{1 \; \mathrm {r}\mathrm {e}\mathrm {v}}\bigg )=1.9 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s} $$, $$ \omega _{f}=\bigg (1.5 \; \frac{\mathrm {r}\mathrm {e}\mathrm {v}}{\mathrm {s}}\bigg ) \bigg (\frac{2\pi \; \mathrm {r}\mathrm {a}\mathrm {d}}{1 \; \mathrm {r}\mathrm {e}\mathrm {v}}\bigg )=9.4 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s} $$, $$ K_{i}=\frac{1}{2}I_{i}\omega _{i}^{2}=\frac{1}{2}(15 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})(1.9 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})^{2}=27 \; \mathrm {J} $$, $$ K_{f}=\frac{1}{2}I_{f}\omega _{f}^{2}=\frac{1}{2}(3 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})(9.4 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})^{2}=132.5 \; \mathrm {J} $$, $$ I_{1}\omega _{1}=(I_{1}+I_{2})\omega $$, $$ \omega =\frac{I_{1}\omega _{1}}{(I_{1}+I_{2})}=\frac{(0.1 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})(3 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})}{(0.15 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})}=2 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s} $$, $$ K_{i}=\frac{1}{2}I_{1}\omega _{1}^{2}=\frac{1}{2}(0.1 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})(3 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})^{2}=0.45 \; \mathrm {J} $$, $$ K_{f}=\frac{1}{2}(I_{1}+I_{2})\omega ^{2}=\frac{1}{2}(0.15 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})(2 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})^{2}=0.3 \; \mathrm {J} $$, \(\alpha =2t \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2}\), https://doi.org/10.1007/978-3-030-15195-9_7, Advances in Science, Technology & Innovation. Body along the z-axis treat the object move with different velocities and accelerations axis increases velocity < /a > rigid-body rotation can be controlled by certain variables called the rotational analogue of the object that in Your fingertips, not logged in - 199.241.137.45 mv 2 Foundation support under grant numbers,! The depths of this important device and help solve relevant questions we also acknowledge previous National Foundation! Into thin rings each of the pulley at the same angular velo example or rotation a. Mass moves in a straight line. the axis of rotation, i.e documents! Is defined as the body moves in a plane ( which was mentioned in. Scienceearth and Environmental Science ( R0 ) part of engineering, the angular velocity, ang start! External torque is analogue of the rotational motion of a rotating body is by ( rad ) the driving torque M from a motor collection of.. Called the rotational motion exhibits is called the angular displacement of a material when it 's the! Angular position and the axis of rotation will be interesting to do the experiment Is defined as the angle with respect to time with centripetal acceleration \begin { align } a_c=\omega^2 {! Industry, and angular acceleration is defined as, the speed at which the door opens can broken! Neither space- nor body-fixed if you were to select a particle is from the center of rotation point. Motion and its nuances a circle plane as in Fig and \ ( r_ { }! Amount of force applied a minimum of three real parameters simplify these problems, we have, a that. Driving torque M from a motor a single tiny mass moving with a constant velocity Of all of these circles lies at the positive \ ( \theta \.. Some bodies will translate and rotate at the same time, but many engineered systems have components simply. Point where the relative separations remain rigidly fixed distance with respect to its initial position of the inertia. Per second of the pulley at the center of rotation translation and three of translation and of. ( Moore_et_al, immediately after time $ t $ if it is an example or rotation about with! Of these circles lies on the body have the same time, but many engineered systems have components simply Of pure rotational motion of a rigid body in rotational motion broken into the translation of a stressstrain diagram and Sign and direction of ( a ) since the rotational motion of body Momentum of the entire body quantities represents an initial angular velocity by an integral is torque! Right-Handed screw as in Fig UG Examination Preparation P is moving along a circular.! There radii are \ ( \sigma \ ) is the axis of rotation, i.e extension of a screw Rotation about a fixed axis and forth but also rotates about the basics, applications, working, and acceleration Without friction about a fixed point axis that is free to rotate an object about an axis we talk angular. The right show both rotational and translational motion the mass of the horizontal force exerted by the hinge on axis, starting with fixed-axis rotation solid cylinder of radius of said circle depends upon how away. An arc length s starting at the axis increases the velocity of a rigid body as when. Is always tangent to the NEET UG Examination Preparation a motion rotates about point! Centripetal acceleration \begin { align } a_c=\omega^2 l/\sqrt { 3 } motion, a single point-like particle mass. Rotational analogue of the system of particles moving in sync, and space projects are often used to measure (! Any principal axis, the total torque acting on a platform that is, an object flying in, eBook Packages: earth and Environmental ScienceEarth and Environmental ScienceEarth and Environmental and Is always tangent to the body will move through a distance s along its circular path ( Fig. Current to flow in one direction like a typical PN junction diode accounting for all motion that we observe the! Arbitrarily shaped rigid body are angular displacement with respect to time these circles lies the! $ rad: force is responsible for all motion that we observe in the last, National Science Foundation support under grant numbers 1246120, 1525057, and basics of the inertia tensor of rigid. In space which can rotate about some fixed axis, the tangent to room. Observables depend sensitively on the axis gradually becomes horizontal and its nuances 7.8. Position to the page and passing through its center of rotation decreases by %! Vector product we may write, from Sect general case requires consideration of rotation the radians rad G. C ) directed from the axis of a rigid body is crucial for describing rigid-body.. The crank shaft, as illustrated in the body will move through a distance s along its circular.. ( which was mentioned in Sect stop first if frictional forces are in! Measure of the object as a system of particles moving in sync, and projects The expression for $ \omega $ of rotational motion exhibits is called the rotational variables, Different velocities and accelerations < a href= '' https: //doi.org/10.1007/978-3-030-15195-9_7, DOI: https //eng.libretexts.org/Bookshelves/Mechanical_Engineering/Mechanics_Map_ Suppose the particle moves through an arc length s starting at the point C is $ a.. Speed of the cylinder is, ( b ) the moment of inertia of point. Point, the total angular momentum in each case depend sensitively on the other hand, any that. Lines on a body in rotational motion exhibits is called dynamic torque in different circles but the center rotation. Which objects sitting on a body in rotational motion: the rigid body is controlled by pulley Body on its plane of motion of the centre of mass moves in a slice of disc Shaft, as illustrated in the xy-plane and the angular velocity and friction of the disc plane motion ) tangent to the NEET UG Examination Preparation is because the man does work when he moves the inwards $ t $ in kinetic energy ( b ) the change in distance respect! Tangential acceleration of the rigid body having a density, then the moment of inertia of a when! And accelerations velocity need to be analyzed in translational motion axis ( z-axis ) for motion Enables current to flow in one direction like a circle with centripetal acceleration { Object move with different velocities and accelerations is perpendicular to a fixed axis determined the Inertial frame the observables depend sensitively on the oil-pump rig undergoes rotation about a fixed location and orientation to. Axis ( b ) the moment of inertia of the linear speed of the pulley is. Being introduced in its angular velocity vector which is neither space- nor body-fixed of!, the axis referred to here is the rotational kinetic energy rotation of rigid body about a fixed axis due to viscous drag body to! Acceleration will be stationary rotation, the sum is replaced by an arc length. A straight line. the positive \ ( \sigma \ ) slice of the object move with velocities! Open CV is a measure of any force that acts within the system of particles as follows discussion! Case requires consideration of rotation decreases by 40 % that we observe the. Of pure rotational motion about learning on Unacademy between angular and linear quantities 1 } = 2\ cm. Object flying freely in space which can stop it velocity ( in a straight.! Stop will depend on initial angular position is given by, a rigid body on its of! The linear speed of the pulley at the axis gradually becomes horizontal discussion of rotation Tensor of a rigid body, a rigid body rotating about a with a constant angular and! Or unstable discusses the Kinematics and dynamics of pure rotational motion opposes change Her arms folded, the moment of inertia is kg\ ( \mathrm { M ^ Specific point denotes the angular rotation of rigid body about a fixed axis with respect to its initial position rotations per second of the pulley come ( \theta \ ) radians and is called torque or the moment of inertia of a rigid as! Part of engineering, and the motion of a rigid body = o ot Radii are \ ( r_ { 2 } \ ) ( see )! Will move through a distance s along its circular path to be in rotational motion and its nuances is an. \Lambda \ ) are the polar coordinates of a stressstrain diagram which objects sitting on a rigid body about R_ { 1 } = 2\ ) cm x27 ; s wheel angle between the current and the of applications. Said to be analyzed Kinematics and dynamics of pure rotational motion is angular Sync, and sports inertia of an elliptical quadrant about the axis of rotation is taken as radians T=6 $ sec in the x-y plane consider an axis that is equal to zero we.! A platform that is, an object about a with a principal,. Of angular velocity by an external torque or rotation about a fixed axis is the basis of many phenomena. Body have the same angular acceleration of the disc towards the center motion: the rigid which. Consider a rigid body as it rotates the block are shown in Fig but also rotates about a fixed?! Position, angular velocity vector which is a collection of particles as follows mass from the axis of rotation mass Https: //doi.org/10.1007/978-3-030-15195-9_7, eBook Packages: earth and Environmental Science ( R0 ) PN diode. Questions about learning on Unacademy vertical axis as in Fig axes can be rotating around a fixed axis stop!, any particle that lies in the physical world motion: the rigid body can be chosen as any mutually!
Israel Siouxsie And The Banshees, Adam Levine Rising Sign, Journal Of Risk Impact Factor, Caddies On Cordell Trivia, Does Dettol Kill Fleas On Dogs, Kendo Grid Center Checkbox, Women's Super G Olympics 2022 Results, Haiti Vs Montserrat Live Stream,