0000005924 00000 n Body-Fixed and Space-Fixed Frames of Reference - Stanford University When a rigid object rotates about a fixed axis, what is true When a rigid body rotates about a fixed axis - Numerade; FAQs. [Physics] Rotation of rigid body with two different angular velocities TR=I_O\alpha=(MR^2/2)\alpha, At \(t=2 \; \mathrm {s}\) Find (a) the angular speed of the wheel (b) the angle in radians through which the wheel rotates (c) the tangential and radial acceleration of a point at the rim of the wheel. Rotation About a Fixed Axis - researchgate.net 7.10 is valid for any rigid object in pure rotation where it only gives the component of the angular momentum that is parallel to the rotational axis. Rotational Motion of a Rigid Body. Dynamics of Rotational Motion about a Fixed Axis - Toppr-guides Now, consider the raw egg. Advances in Science, Technology & Innovation. Obtain the x-component and the y-component of the force exerted by the hinge on the body, immediately after time $T$. 0000001867 00000 n Different particles move in different circles but the center of these circles lies at the axis of rotation. Consider the three masses and the connecting rods together as a system. (PDF) On the rotational motion of a rigid body - ResearchGate \begin{align} These two accelerations should be equal for no slip at C i.e., CHAPTER 5 PLANAR KINEMATICS OF A RIGID BODY - SlideToDoc.com Solution: This particle (at point P) will rotate in a circle of fixed radius r which represents the perpendicular distance from \(\mathrm {P}\) to the axis of rotation. the z-axis) by lz, then lz = CP vector mv vector = m(rperpendicular)^2 k cap and l = lz + OC vector mv vector We note that lz is parallel to the fixed axis, but l is not. If the moment of inertia of the pulley about its axis of rotation is 10 kg-m2, then the number of rotations made by the pulley before its direction of motion is reversed is, Solution: 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. 0000009574 00000 n 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. Rotation of a Rigid Body About a Fixed Point - L1 | Definition Thus, to find the rotational inertia, the axis of rotation must be specified. The tangential acceleration of the pulley at the point C is $\alpha R$. A wheel of mass of 20 kg and radius of 0.75 \(\mathrm {m}\) is initially rotating at 120 rev/min. \begin{align} The motion of the object is contained in the xy-plane and the axis of rotation is along the z-axis. Now consider another axis that is parallel to the first axis and that passes through a point \(\mathrm {P}\) as shown in Fig. The axis referred to here is the rotation axis of the tensor . Draw a free body diagram accounting for all external forces and couples. This follows from Eq. C) m aG D) m aO. 7.2 shows analogous equations in linear motion and rotational motion about a fixed axis. The prior discussion in chapter \((2.12)\) showed that rigid-body rotation is more complicated than assumed in introductory treatments of rigid-body rotation. 0000000961 00000 n \begin{align} The centre of mass of the system (G) is at a distance $\mathrm{AG}={l}/{\sqrt{3}}$ from the hinge point A. The angular momentum of the rigid body about the fixed axis of rotation is given by 1. A man stands on a platform that is free to rotate without friction about a vertical axis as in Fig. rigid body does not exist, it is a useful idealization. If a raw egg and a boiled egg are spinned together with same angular velocity on the horizontal surface then which one will stops first? In rotation of a rigid body about a fixed axis, every ___A___ of the body moves in a ___B___, which lies in a plane ___C___ to the axis and has its centre on the axis. Slideshow 3144668 by mina (c) After seven revolutions the angular velocity is, that gives \(\omega =16.24 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}\). The angular Momentum of a Rigid Object Rotating and Translating. The speed at which the door opens can be controlled by the amount of force applied. Mechanics Map - Fixed Axis Rotation - Pennsylvania State University Since rotation here is about a fixed axis, every particle constituting the rigid body behaves to be rotating around a fixed axis. 0000004127 00000 n Find (in vector form) the linear velocity and acceleration of the point \(\mathrm {P}\) on the bar. \begin{align} When rotating about a fixed axis, a rigid body constantly changes its angle with respect to its initial position and the fixed axis. Hence, the instantaneous angular velocity and acceleration (\(\omega \) and \(\alpha \)) can be represented by vectors but not their average values (\(\overline{\omega }\) and \(\overline{\alpha }\)). https://doi.org/10.1007/978-3-030-15195-9_7, DOI: https://doi.org/10.1007/978-3-030-15195-9_7, eBook Packages: Earth and Environmental ScienceEarth and Environmental Science (R0). A man stands on a platform that is free to rotate without friction about a vertical axis, Because the resultant external torque on the system is zero, it follows that the total angular momentum of the system is conserved. 0000006467 00000 n Recall d d = or dt = dt d d 2 d = = = dt dt 2 d Uniform Rotation (angular acceleration=0 ) = 0 +t Uniformly Accelerated Rotation( angular acceleration = constant): . A ballet dancer spins about a vertical axis 120 rpm with arms outstretched. The discussion of general rotation, in which both the position and the direction of the axis change, is quite complex. The parameters that govern the rotational motion of a rigid body are angular displacement, angular velocity, and angular acceleration. Jul 22, 2009 #3 Amar.alchemy 79 0 cepheid said: Correspondence to Abstract. about that axis. 21.2 Translational Equation of Motion . \begin{align} Fixed-axis rotation describes the rotation around a fixed axis of a rigid body; that is, an object that does not . 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). The measure of the change in angular velocity with respect to the time of a rigid body in rotational motion due to the application of an external torque is called angular acceleration. 7.28. For example, when observed in the stationary fixed frame, rapid rotation of a long thin cylindrical pencil about the longitudinal symmetry axis gives a time-averaged shape of the pencil that looks like a thin cylinder, whereas the time-averaged shape is a flat disk for rotation about an axis perpendicular to the symmetry axis of the pencil. 0000004937 00000 n If the angular position of a point on a rotating wheel is given by \(\theta =2t+ 5t^{2}\) rad, find the angular speed and angular acceleration of the point at \(t=2 \; \mathrm {s}.\). Substitute $\vec{a}$ from the previous equation into the last equation to get $F_x=-F/4$ and $F_y=\sqrt{3}m\omega^2l$. PDF Lecture 10 Rotation of a Rigid Object About a Fixed Axis If the angular velocity of the smaller sprocket is 2 \(\mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s},\) find the angular velocity of the other. The disc rotates about a fixed point O. ROTATION ABOUT A FIXED AXIS.pptx - TYPES OF PLANE MOTION, However, for the general case of free rotation, the vector of angular velocity . This is followed by a discussion of practical applications. Find the angular speed of the disc when the man is at a distance of 0.7 \(\mathrm {m}\) from the center if its angular speed when the man starts walking is 1.6 \(\mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}.\), An L-shaped bar rotating counterclockwise, Four masses connected by light rigid rods, A uniform rod of length L and mass M is pivoted at \(\mathrm {O}\). To understand the equilibrium of an extended object. The direction of the linear speed of the particles is always tangent to the path (as mentioned in Sect. 0000002460 00000 n Problem. Module 3 -- Angular Momentum of a Rigid Body both Rotating and \end{align} 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. and its angular acceleration is Equations7.77.9 are the vector relationship between angular and linear quantities. 12.1 Rotational Motion 12.2 Center of Mass 12.3 Rotational energy 12.4 Moment of Inertia 12.5 Torque 12.6 Rotational dynamics 12.7 Rotation about a fixed axis 12.8 *Rigid-body equilibrium 12.9 Rolling Motion. Let \(t_{1}=0, t_{2}=t, \omega _{1}=\omega _{\mathrm {o}}, \omega _{2}=\omega , \theta _{1}=\theta _{\mathrm {o}}\), and \(\theta _{2}=\theta .\) Because the angular acceleration is constant it follows that the angular velocity changes linearly with time and the average angular velocity is given by, Finally solving for t from Eq. 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. As the rigid body rotates, a particle in the body will move through a distance s along its circular path (see Fig. Fixed Rotation of a Rigid Body . The further a particle is from the axis of rotation, the greater the angular velocity and acceleration will be. The motion of electrons about an atom and the motion of the moon about the earth are examples of rotational motion. A light rope wrapped around a uniform cylindrical shell, (a) Because the line of action of both the weight and the normal forces passes through the central axis of the cylinder, they produce no torque. where I is the moment of inertia of the rigid body about the rotational axis (z-axis). Rotation of a rigid body about external axis | Physics Forums This body is placed on a horizontal frictionless table x-y plane) and is hinged to it at the point A, so that it can move without friction about the vertical axis through A (see figur). We talk about angular position, angular velocity, ang. Here, A, B, and C refer to: (a) particle, perpendicular, and circle (b) circle, particle, and perpendicular (c) particle, circle, and perpendicular (d) particle, perpendicular, and perpendicular. We encourage you to find the number of rotations made by the pulley till $t=8$ sec. the person answering this question offers the following expression for the kinetic energy of an object that is translating and rotating about a fixed point: t = 1 2mv2 + 1 22in + mr cm (v ) however, as they present this equation without any sort of proof or derivation or a reference of some kind it is difficult for me to feel assured A uniform disc of moment of inertia of 0.1 kg m\(^{2}\) is rotating without friction with an angular speed of 3 \(\mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}\) about an axle passing through its center of mass as in Fig. In the body-fixed frame, the ``vertical'' axis coincides with the top's axis of rotation (spin). Three particles A, B and C, each of mass $m$, are connected to each other by three massless rigid rods to form a rigid equilateral triangular body of side $l$. There are two types of plane motion, which are given as follows: 1. This is because the finite angular displacement \(\triangle \theta \) does not obey the commutative law of vector addition (see Fig. 7.25. - 199.241.137.45. Alrasheed, S. (2019). \((I_{i}=15 \; \mathrm {kg\, m^2}\) And \(I_{f}=3 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})\). But we must first understand rotational motion and its nuances. It is named after Thomas Young. A projectile of mass m moving at velocity v collides with the rod and sticks to it, You can also search for this author in \label{dic:eqn:3} Since the different points in the body will have different values of r, the value of tangential velocity is not the same for all the points in the body. We treat the whole system as a single point-like particle of mass m located at the center . 7.26 shows Atwoods machine when the mass of the pulley is considered. 0000019769 00000 n Hb```L[(1AaY2C&_TlEC#qf!R[-i1pm7LqSrRUnB3N(\aflFYu +eNS-S519[-H9]iO((tfh`T6 9]::@4R!(M! 28A1_absolute motions.png - RIGID-BODY MOTION: FIXED AXIS ROTATION V Calculating the moment of inertia of a uniform solid cylinder with the volume element defined in different ways, Method 1: Using a single integration by dividing the cylinder into thin cylindrical shells each of radius r, length L and thickness dr as in Fig. 7.7), thus, The instantaneous power delivered to rotate an object about a fixed axis is found from, Table. 0000001474 00000 n where $I_O$ is moment of inertia of the uniform solid disc about the axis of rotation. 0000005734 00000 n Ans : Angular velocity is the rate of change in angular displacement with respect to time. $$ 1 \; \text {rev} =360^{\circ }=2\pi \; \text {rad} $$, $$ 1 \; \text {rad} =57.3^{\circ }=0.159 \; \text {rev} $$, \(2\pi \mathrm {r}\mathrm {a}\mathrm {d}\), $$ \theta =(2\pi +2\pi +2\pi ) \; \text {rad} =6\pi \; \text {rad} $$, $$ \triangle \theta =\theta _{2}-\theta _{1} $$, $$ \overline{\omega }=\frac{\theta _{2}-\theta _{1}}{t_{2}-t_{1}}=\frac{\triangle \theta }{\triangle t} $$, $$ \omega =\lim _{\triangle t\rightarrow 0}\frac{\triangle \theta }{\triangle t}=\frac{d\theta }{dt} $$, \(\mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}\), $$ \overline{\alpha }=\frac{\omega _{2}-\omega _{1}}{t_{2}-t_{1}}=\frac{\triangle \omega }{\triangle t} $$, $$ \alpha =\lim _{\triangle t\rightarrow 0}\frac{\triangle \omega }{\triangle t}=\frac{d\omega }{dt} $$, \(\mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2}\), \(\mathrm {r}\mathrm {e}\mathrm {v}/\mathrm {s}^{2}\), \(\mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}.\), $$ 15^{\mathrm {o}}=(15 \; \displaystyle \mathrm {deg})\bigg (\frac{1 \; \mathrm {r}\mathrm {e}\mathrm {v}}{360 \; \mathrm {deg}}\bigg )=0.042 \; \text {rev} $$, $$ 15^{\mathrm {o}}=(15 \; \displaystyle \mathrm {deg})\bigg (\frac{2 \; \pi \mathrm {r}\mathrm {a}\mathrm {d}}{360 \; \mathrm {deg}}\bigg )=0.26 \; \text {rad} $$, $$ 0.25 \; \displaystyle \mathrm {r}\mathrm {e}\mathrm {v}/\mathrm {s}^{2}=\bigg (0.25 \; \frac{\mathrm {r}\mathrm {e}\mathrm {v}}{\mathrm {s}^{2}}\bigg )\bigg (\frac{2\pi \; \mathrm {r}\mathrm {a}\mathrm {d}}{1\mathrm {r}\mathrm {e}\mathrm {v}}\bigg )=1.57 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2} $$, $$ 0.25 \; \displaystyle \mathrm {r}\mathrm {e}\mathrm {v}/\mathrm {s}^{2}=\bigg (0.25 \; \frac{\mathrm {r}\mathrm {e}\mathrm {v}}{\mathrm {s}^{2}}\bigg )\bigg (\frac{360 \; \mathrm {deg}}{1\,\mathrm {r}\mathrm {e}\mathrm {v}}\bigg )=90 \; \mathrm {deg}/\mathrm {s}^{2} $$, $$ 3\ \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}=\bigg (3 \; \frac{\mathrm {r}\mathrm {a}\mathrm {d}}{\mathrm {s}}\bigg )\bigg (\frac{1 \; \mathrm {r}\mathrm {e}\mathrm {v}}{2\pi \; \mathrm {r}\mathrm {a}\mathrm {d}}\bigg )=0.48 \; \mathrm {r}\mathrm {e}\mathrm {v}/\mathrm {s} $$, $$ 3 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}=\bigg (3 \; \frac{\mathrm {r}\mathrm {a}\mathrm {d}}{\mathrm {s}}\bigg )\bigg (\frac{360^{\mathrm {o}} \; \mathrm {deg}}{2\pi \; \mathrm {r}\mathrm {a}\mathrm {d}}\bigg )=172 \; \mathrm {deg}/\mathrm {s} $$, $$ \theta _{1}=((0.3)(1 \; \mathrm {s})^{2}+(0.4)(1 \; \mathrm {s})^{3})=0.7 \; \text {rad} $$, $$ \theta _{2}=((0.3)(2 \; \mathrm {s})^{2}+(0.4)(2 \; \mathrm {s})^{3})=4.4 \; \text {rad} $$, $$ \triangle \theta =( 4.4 \; \text {rad})-(0.7\,\text {rad}) =3.7 \; \text {rad} $$, $$ \overline{\omega }=\frac{\triangle \theta }{\triangle t}=\frac{(3.7 \; \mathrm {r}\mathrm {a}\mathrm {d})}{(1 \; \mathrm {s})}=3.7 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s} $$, $$ \omega =\frac{d\theta }{dt}=((0.6)t+(1.2)t^{2}) \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s} $$, $$ \omega =(0.6)(5 \; \mathrm {s})+(1.2)(5 \; \mathrm {s})^{2}=33 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s} $$, $$ \alpha =\frac{d\omega }{dt}=((0.6)+(2.4)t) \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2} $$, $$ \alpha =(0.6)+(2.4)(5 \; \mathrm {s})=12.6 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2} $$, \(\alpha =(9-2t) \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2}\), $$ \omega =\int \alpha dt=\int (9-2t)dt=9t-t^{2}+c_{1} $$, \(\omega =2 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}\), \(c_{1}=2 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}\), $$ \omega =(9t-t^{2}+2) \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s} $$, $$ \theta =\int \omega dt=\int (9t-t^{2}+2)dt=\frac{9}{2}t^{2}-\frac{1}{3}t^{3}+2t+c_{2} $$, \(t=0, \theta =3 \; \mathrm {r}\mathrm {a}\mathrm {d}\), $$ \displaystyle \theta =\bigg (\frac{9}{2}t^{2}-\frac{1}{3}t^{3}+2t+3 \bigg ) \; \text {rad} $$, $$ \displaystyle \theta =\frac{9}{2}(4.5 \; \mathrm {s})^{2}-\frac{1}{3}(4.5 \; \mathrm {s})^{3}+2(4.5 \; \mathrm {s})+3=72.8 \; \text {rad} $$, \(t_{1}=0, t_{2}=t, \omega _{1}=\omega _{\mathrm {o}}, \omega _{2}=\omega , \theta _{1}=\theta _{\mathrm {o}}\), $$ \overline{\omega }=\frac{\omega _{0}+\omega }{2} $$, $$ \alpha =\overline{\alpha }=\frac{\omega _{2}-\omega _{1}}{t_{2}-t_{1}}=\frac{\omega -\omega _{0}}{t} $$, $$\begin{aligned} \omega =\omega _{0}+\alpha t \end{aligned}$$, $$ \overline{\omega }=\frac{\theta _{2}-\theta _{1}}{t_{2}-t_{1}}=\frac{\theta -\theta _{0}}{t}=\frac{\omega _{0}+\omega }{2} $$, $$\begin{aligned} \displaystyle \theta =\theta _{0}+\frac{1}{2}(\omega _{0}+\omega )t \end{aligned}$$, $$ \theta =\theta _{0}+\frac{1}{2}(\omega _{0}+\omega )t=\theta _{0}+\frac{1}{2}(\omega _{0}+\omega _{0}+\alpha t)t $$, $$\begin{aligned} \displaystyle \theta =\theta _{0}+\omega _{0}t+\frac{1}{2}\alpha t^{2} \end{aligned}$$, $$ \theta =\theta _{0}+\frac{1}{2}(\omega _{0}+\omega )t=\theta _{0}+\frac{1}{2}(\omega _{0}+\omega )\left( \frac{\omega -\omega _{0}}{\alpha }\right) $$, $$\begin{aligned} \omega ^{2}=\omega _{0}^{2}+2\alpha (\theta -\theta _{0}) \end{aligned}$$, $$\begin{aligned} v=r\omega \end{aligned}$$, $$\begin{aligned} a_{r}=\displaystyle \frac{v^{2}}{r} \end{aligned}$$, $$ a_{t}=\frac{dv}{dt}=r\frac{d\omega }{dt} $$, $$ \mathbf {a}=\mathbf {a}_t+\mathbf {a}_r $$, $$ a=\sqrt{{a_t}^2+{a_r}^2}=\sqrt{{r}^2{\alpha }^2+{r}^2{\omega }^4}=r\sqrt{{\alpha }^2+{\omega }^4} $$, \(\theta =(60\;\mathrm {deg})(2\pi \mathrm {r}\mathrm {a}\mathrm {d}/360\;\mathrm {deg})=1.05\), $$ \theta =\theta _{0}+\omega _{0}t+\frac{1}{2}\alpha t^{2} $$, $$ \alpha =\frac{2\theta }{t^{2}}=\frac{2(1.05 \; \mathrm {r}\mathrm {a}\mathrm {d})}{(2 \; \mathrm {s})^{2}}=0.525 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2} $$, $$ \omega =\omega _{0}+\alpha t=(0.525 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2})(2 \; \mathrm {s})=1.05 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s} $$, $$ \omega =(0.525 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2})(6\mathrm {s})=3.15 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s} $$, $$ v=r\omega =(0.07 \; \mathrm {m})(1.05 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})=0.074 \; \mathrm {m}/\mathrm {s} $$, $$ s=r\theta =(0.07 \; \mathrm {m})(1.05 \; \mathrm {rad} ) =0.074 \; \mathrm {m} $$, \(\mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s},\), $$ r_{1}\omega _{1}=r_{2}\omega _{2}=v $$, $$ \omega _{2}=\frac{r_{1}}{r_{2}}\omega _{1}=\frac{(2 \; \mathrm {c}\mathrm {m})}{(5 \; \mathrm {c}\mathrm {m})}(2 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})=0.8 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s} $$, $$ \displaystyle \omega =rv=\frac{2\pi }{T}=\frac{2(3.14)}{(27.3 \; \mathrm {d}\mathrm {a}\mathrm {y})}=0.23 \; \text {rad/day} $$, $$ \displaystyle \omega = \bigg (0.23 \; \frac{\mathrm {r}\mathrm {a}\mathrm {d}}{\mathrm {d}\mathrm {a}\mathrm {y}}\bigg ) \bigg (\frac{1 \; \mathrm {r}\mathrm {e}\mathrm {v}}{2\pi \; \mathrm {r}\mathrm {a}\mathrm {d}}\bigg )=0.037 \; \text {rev/day} $$, $$\begin{aligned} \mathbf {v}=\varvec{\omega }\times \mathbf {R} \end{aligned}$$, $$ \mathbf {a}=\frac{d\mathbf {v}}{dt}=\frac{d}{dt}(\varvec{\omega }\times \mathbf {R}) $$, \((d/dt(\mathbf {A}\times \mathbf {B})=\mathbf {A}\times d\mathbf {B}/dt+d\mathbf {A}/dt\times \mathbf {B})\), $$ \mathbf {a}=\frac{d \varvec{\omega }}{dt}\times \mathbf {R}+\varvec{\omega }\times \frac{d\mathbf {R}}{dt} $$, $$ =\varvec{\alpha }\times \mathbf {R}+\varvec{\omega }\times \mathbf {v} $$, $$ |\varvec{\alpha }\times \mathbf {R}|=\alpha R\sin \theta =r\alpha =a_{t} $$, $$\begin{aligned} \mathbf {a_{t}}=\varvec{\alpha }\times \mathbf {R} \end{aligned}$$, $$ |\varvec{\omega }\times \mathbf {v}|=\omega v\sin 90^{\mathrm {o}}=\omega v=r\omega ^{2}=a_{r} $$, $$\begin{aligned} \mathbf {a}_{r}=\varvec{\omega }\times \mathbf {v} \end{aligned}$$, \(K=\displaystyle \frac{1}{2}\sum _{i}m_{i}v_{i}^{2}\), $$ K_{R}=\frac{1}{2}\sum _{i}(m_{i}r_{i}^{2})\omega ^{2} $$, $$ I=\lim _{\triangle m_{\mathrm {i}\rightarrow 0}}\sum _{i}m_{i}r_{i}^{2}=\int r^{2}dm $$, $$ I_{cm}=\int r^{2}dm=\int (x^{2}+y^{2})dm $$, $$ I_{P}=\int [(x-x_{P})^{2}+(y-y_{P})^{2}]dm $$, $$ I_{P}=\int (x^{2}+y^{2})dm-2x_{P}\int xdm-2y_{P}\int ydm+\int (x_{P}^{2}+y_{P}^{2})dm $$, $$ I_{P}=I_{cm}+MD^{2} \quad \text {(Parallel--Axis Theorem)} $$, $$ \mathbf {L}_{i}=\mathbf {R}_{i}\times \mathbf {p}_{i} $$, $$ L_{iz}=L_{i}\sin \theta =R_{i}p_{i}\sin \theta =R_{i} ({ m_{i} v_{i}}) \sin \theta $$, $$ =R_{i}m_{i}(r_{i}\omega )\sin \theta =m_{i}r_{i}^{2}\omega $$, $$ L_{z}=\sum _{i}m_{i}r_{i}^{2}\omega =\bigg (\sum _{i}m_{i}r_{i}^{2}\bigg )\omega $$, $$\begin{aligned} \mathbf {L}_{z}=I\varvec{\omega } \end{aligned}$$, $$ \mathbf {L}=\sum _{i}\mathbf {L}_{iz}=\mathbf {L}_{z}=I\varvec{\omega }$$, $$\begin{aligned} \mathbf {L}=I\varvec{\omega } \end{aligned}$$, $$ \Sigma {\varvec{\tau }_{ext}}=\frac{d\mathbf {L}}{dt} $$, $$ \Sigma {\varvec{\tau }_{extz}}=\frac{d\mathbf {L}_{z}}{dt}=\frac{d(I\varvec{\omega })}{dt}=I\varvec{\alpha }$$, $$ \Sigma {\varvec{\tau }_{ext}}=\frac{d\mathbf {L}}{dt}=\frac{d(I\varvec{\omega })}{dt}=I\varvec{\alpha }$$, $$ \triangle \theta =( 10 \; \mathrm {rev}) \bigg (\displaystyle \frac{2\pi \; \mathrm {r}\mathrm {a}\mathrm {d}}{1 \; \mathrm {r}\mathrm {e}\mathrm {v}}\bigg )=62.8 \; \text {rad} $$, $$ \alpha =\frac{\omega ^{2}-\omega _{0}^{2}}{2\triangle \theta }=\frac{0-(5 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})^{2}}{2(62.8 \; \mathrm {r}\mathrm {a}\mathrm {d})}=-0.2 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s} $$, $$ \tau =I\alpha =MR^{2}\alpha =(5 \; \mathrm {k}\mathrm {g})(0.1 \; \mathrm {m})^{2}(-0.2 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2})=-0.01 \; \mathrm {N}\,\mathrm {m} $$, $$\begin{aligned} I_{z}&=\displaystyle \sum _{i}m_{i}r_{i}^{2}=2ma^{2}+\frac{m}{2}a^{2}+ma^{2}=\frac{7}{2}ma^{2}\\&=\frac{7}{2}(0.1 \; \mathrm {k}\mathrm {g})(0.2 \; \mathrm {m})^{2}=0.014 \; \mathrm {kg\, m^2} \end{aligned}$$, $$ K_{R}=\frac{1}{2}I_{z}\omega ^{2}=\frac{1}{2}(0.014 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})(5 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})^{2}=0.175 \; \mathrm {J} $$, $$I_{y}=\displaystyle \frac{m}{2}a^{2}+2ma^{2}=\frac{5}{2}ma^{2}=\frac{5}{2}(0.1 \; \mathrm {k}\mathrm {g})(0.2 \; \mathrm {m})^{2}=0.01 \; \mathrm {kg\, m^2}$$, $$ K_{R}=\frac{1}{2}I_{y}\omega ^{2}=\frac{1}{2}(0.01 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})(5 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})^{2}=0.125 \; \mathrm {J} $$, $$ I_{x}=ma^{2}=(0.1 \; \mathrm {k}\mathrm {g})(0.2 \; \mathrm {m})^{2}=4\times 10^{-3} \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2} $$, $$ K_{R}=\frac{1}{2}I_{x}\omega ^{2}=\frac{1}{2}(4\times 10^{-3} \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})(5 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})^{2}=0.05 \; \mathrm {J} $$, $$ dm=\lambda dx=\bigg (\frac{M}{L}\bigg )dx $$, $$ I_{cm}=I_{y}=\displaystyle \int r^{2}dm=\int _{x=-\frac{L}{2}}^{\frac{L}{2}}x^{2}\bigg (\frac{M}{L}\bigg )dx=\frac{M}{L}\bigg (\frac{x^{3}}{3}\bigg ) \bigg |_{-L/2}^{L/2}=\displaystyle \frac{1}{12}ML^{2} $$, $$ I_{y'}=I_{cm}+MD^{2}=\frac{1}{12}ML^{2}+M\bigg (\frac{L}{2}\bigg )^{2}=\frac{1}{3}ML^{2} $$, $$ I_{y''}=I_{cm}+MD^{2}=\frac{1}{12}ML^{2}+M\bigg (\frac{L}{2}-\frac{L}{6}\bigg )^{2}=\frac{7}{36}ML^{2} $$, $$ I_{cm}=\int r^{2}dm=\int r^{2}\sigma dA=\int _{y=-a/2}^{a/2}\int _{y=-b/2}^{b/2}{(x^{2}+y^{2})\bigg (\frac{M}{ab}\bigg )dxdy} $$, $$ =\displaystyle \frac{M}{ab} \int _{y=-a/2}^{a/2} {{\bigg (\frac{x^3}{3}+xy^2\bigg )|_{x=-b/2}^{b/2}}dy}=\frac{M}{ab} \int _{y=-a/2}^{a/2} {{\bigg (\frac{b^3}{12}+by^2\bigg )}dy} $$, $$ =\displaystyle \frac{M}{ab} \bigg (\frac{b^3y}{12}+\frac{y^3b}{3}\bigg ) \bigg |_{x=-a/2}^{a/2}=\frac{M}{ab} \bigg [\frac{ab^3}{12}+\frac{ab^3}{12}\bigg ]=\frac{1}{12}M\big (a^2+b^2\bigg ) $$, $$ I=\int r^{2}dm=\int _{0}^{R}r^{2}(\rho 2\pi rLdr)=2\pi \rho L\int _{0}^{R}r^{3}dr=\frac{\pi \rho L}{2}R^{4} $$, $$ I=\int r^{2}dm=\int _{0}^{2\pi } \int _{r=0}^{R} r^{3}\rho Ldrd\theta =\rho \frac{L}{4}R^{4}\int _{\theta =0}^{2\pi } d\theta =\frac{\pi \rho LR^{4}}{2} $$, $$ I=\int r^{2}dm= \int _{\theta =0}^{2\pi } \int _{r=0}^{R} \int _{z=0}^{L}\rho r^{3}drd\theta dz=\rho L\frac{R^{4}}{4}\int _{\theta =0}^{2\pi } d\theta =\frac{\pi \rho LR^{4}}{2} $$, $$ I=I_{1}+I_{2}+I_{3}=3\bigg (\frac{1}{3}ML^{2}\bigg )=ML^{2} $$, $$ dA=2\pi R\sin \theta Rd\theta =2\pi R^{2}\sin \theta d\theta $$, $$ I=\int r^{2}dm=\int R^{2}\sin ^{2}\theta \sigma 2\pi R^{2}\sin \theta d\theta $$, $$ I=\frac{M}{2}R^{2}\int _{\theta =0}^{\pi }\sin ^{3}\theta d\theta =\frac{M}{2}R^{2}\int _{\theta =0}^{\pi }(1-\cos ^{2}\theta )\sin \theta d\theta $$, $$ =\frac{M}{2}R^{2}\bigg [-\cos \theta +\frac{\cos ^{3}\theta }{3}\bigg ]_{\theta =0}^{\pi }=\frac{2}{3}MR^{2} $$, $$ \mathbf {L}_{i}=\mathbf {L}_{f}= \mathrm{constant~(isolated~system)} $$, $$ dW=\mathbf {F}\cdot d\mathbf {s}=\mathbf {F}\cdot \frac{d\mathbf {s}}{dt}dt=\mathbf {F}\cdot \mathbf {v} dt=\mathbf {F}\cdot (\varvec{\omega }\times \mathbf {r})dt $$, $$ =(\mathbf {r}\times \mathbf {F})\cdot \varvec{\omega }dt=\varvec{\tau }\cdot \varvec{\omega }dt $$, $$ dW=\tau \omega dt=\tau \frac{d\theta }{dt}dt=\tau d\theta $$, $$\begin{aligned} W=\displaystyle \int _{\theta _{1}}^{\theta _{2}}\tau d\theta \end{aligned}$$, $$ W=\tau (\theta _{2}-\theta _{1})=\tau \triangle \theta $$, $$ W=\int _{\theta _{1}}^{\theta _{2}}\tau d\theta =\int _{\theta _{1}}^{\theta _{2}}I\alpha d\theta =\int _{\omega _{1}}^{\omega _{2}}I\omega \frac{d\omega }{dt}dt=\int _{\omega _{1}}^{\omega _{2}}I\omega d\omega =\frac{1}{2}I\omega _{2}^{2}-\frac{1}{2}I\omega _{1}^{2} $$, $$ W=\triangle K=\frac{1}{2}I\omega _{2}^{2}-\frac{1}{2}I\omega _{1}^{2} $$, $$ P=\frac{dW}{dt}=\frac{\tau _{z}d\theta }{dt}=\tau _{z}\omega _{z} $$, $$ I=\displaystyle \frac{1}{2}MR^{2}=\frac{1}{2}(5 \; \mathrm {k}\mathrm {g})(0.08 \; \mathrm {m})^{2}=0.016 \; \mathrm {kg\, m^2} $$, $$ \omega =\bigg (\frac{170 \; \mathrm {r}\mathrm {e}\mathrm {v}}{\min }\bigg )\bigg (\frac{2\pi \; \mathrm {r}\mathrm {a}\mathrm {d}}{1 \; \mathrm {r}\mathrm {e}\mathrm {v}}\bigg )\bigg (\frac{1 \; \min }{60 \; \mathrm {s}}\bigg )=17.8 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s} $$, $$ K=\frac{1}{2}I\omega ^{2}=\frac{1}{2}(0.016 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})(17.8 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})^{2}=2.5 \; \mathrm {J} $$, $$ P=(0.2 \; \mathrm {h}\mathrm {p}\ ) \bigg (\frac{746 \; \mathrm {W}}{1\mathrm {h}\mathrm {p}}\bigg )=149.2 \; \mathrm {W} $$, $$ \tau =\frac{P}{\omega }=\frac{(149.2 \; \mathrm {W})}{(17.8 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})}=8.4 \; \mathrm {N}\,\mathrm {m} $$, \(\mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}, (\mathrm {e})\), $$ \tau =FR=(35 \; \mathrm {N})(0.2 \; \mathrm {m})=7 \; \text {N/m} $$, $$ I=MR^{2}=(30 \; \mathrm {k}\mathrm {g})(0.2 \; \mathrm {m})^{2}=1.2 \; \mathrm {kg\, m^2} $$, $$ \alpha =\frac{\tau }{I}=\frac{(7 \; \mathrm {N}\,\mathrm {m})}{(1.2\;\mathrm {k}\mathrm {g}\,\mathrm {m}^{2})}=5.8 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2} $$, $$ a=R\alpha =(0.2 \; \mathrm {m})(5.8 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2})=1.2 \; \mathrm {m}/\mathrm {s}^{2} $$, $$ \omega ^{2}=\omega _{0}^{2}+2\alpha \theta $$, $$ \displaystyle \theta =\frac{(12\,\mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})^{2}}{2(5.8\,\mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2})}=12.4 \; \mathrm {rad} $$, $$ \theta =(12.4 \; \mathrm {rad}) \bigg (\displaystyle \frac{1 \; \mathrm {r}\mathrm {e}\mathrm {v}}{2\pi \; \mathrm {r}\mathrm {a}\mathrm {d}}\bigg )= 2 \; \text {rev} $$, $$ \displaystyle \theta =\frac{s}{R}=\frac{(1 \; \mathrm {m})}{(0.2 \; \mathrm {m})}=5 \; \mathrm {rad} $$, $$ W=\int _{\theta _{0}}^{\theta }\tau d\theta =\tau (\theta -\theta _{0})=(7 \; \mathrm {N}\,\mathrm {m})\ ((5 \; \mathrm {r}\mathrm {a}\mathrm {d})-0)=35 \; \mathrm {J} $$, $$ \omega ^{2}=\omega _{0}^{2}+2\alpha \theta =0+2(5.8 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2}) ( 5 \; \mathrm {rad}) $$, \(\omega =7.6 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}\), $$ W=\triangle K=\frac{1}{2}I\omega ^{2}-\frac{1}{2}I\omega _{0}^{2}=\frac{1}{2}(1.2 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})(7.6 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})^{2}-0=35 \; \mathrm {J} $$, $$\displaystyle \tau =\frac{MgL}{2}\cos \theta =\frac{(0.75\; \mathrm {k}\mathrm {g})(9.8 \; \mathrm {m}/\mathrm {s}^{2})(1 \; \mathrm {m})}{2}\cos 30^{\circ }=3.2\,\mathrm {N\, m}$$, $$ I=\frac{1}{3}ML^{2}=\frac{(0.75\;\mathrm {k}\mathrm {g})(1 \; \mathrm {m})^{2}}{3}=0.25 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2} $$, $$ \alpha =\frac{\tau }{I}=\frac{(3.2 \; \mathrm {N}\,\mathrm {m})}{(0.25 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})}=12.8 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2} $$, $$ a_{t}=r\alpha =L\alpha =(1 \; \mathrm {m})(12.8 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2})=12.8 \; \mathrm {m}/\mathrm {s}^{2} $$, $$ 0+Mg\frac{L}{2}(\sin \theta +1)=\frac{1}{2}I\omega ^{2}+0 $$, $$\omega =\sqrt{Mg\frac{L}{I}(\sin \theta +1)}=\sqrt{\frac{(0.75 \; \mathrm {k}\mathrm {g})(9.8 \; \mathrm {m}/\mathrm {s}^{2})(1\mathrm {m})}{(0.25 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})}(\sin 30^{\circ }+1)}=6.64 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}$$, \(F_{1}=10 \; \mathrm {N}, F_{2}=20 \; \mathrm {N}\), $$\begin{aligned} \tau _{\mathrm {n}\mathrm {e}\mathrm {t}}&=\tau _{1}+\tau _{2}+\tau _{3}=(10 \; \mathrm {N})(0.05\;\mathrm {m})+(20 \; \mathrm {N})(0.05 \; \mathrm {m})\\&-(15 \; \mathrm {N})(0.15 \; \mathrm {m})=-0.75 \; \mathrm {N\, m} \end{aligned}$$, $$\begin{aligned} a=\frac{mg-T}{m} \end{aligned}$$, $$\begin{aligned} a=R\displaystyle \alpha =\frac{TR^{2}}{I} \end{aligned}$$, $$ T=\frac{g}{1/m+R^{2}/I}=\frac{g}{1/m+2R^{2}/MR^{2}} $$, $$ a=\frac{TR^{2}}{I}=\frac{2TR^{2}}{MR^{2}} $$, $$ \alpha =\frac{a}{R}=\frac{g}{R(1+M/{2m})} $$, $$\displaystyle \tau =I\alpha =\frac{2}{5}MR^{2}\alpha =\frac{2}{5}(4.7 \; \mathrm {k}\mathrm {g})(0.05 \; \mathrm {m})^{2}(3\,\mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2})=0.014 \; \mathrm {N\,} $$, $$ \theta =( 7 \; \mathrm {rev}) \bigg (\displaystyle \frac{2\pi \mathrm {r}\mathrm {a}\mathrm {d}}{1 \; \mathrm {r}\mathrm {e}\mathrm {v}}\bigg )=44 \; \text{ rad } $$, $$ W=\tau \triangle \theta = (0.014 \; \mathrm {N/m}) (44 \; \mathrm {rad}) =0.6 \; \mathrm {J} $$, $$ \omega ^{2}=\omega _{0}^{2}+2\alpha (\theta -\theta _{0}) $$, $$ \omega ^{2}=2\alpha \theta =2(3 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2}) ( 44 \; \text {rad}) $$, \(\omega =16.24 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}\), $$ W=\frac{1}{2}I\omega ^{2}-\frac{1}{2}I\omega _{0}^{2}=\frac{1}{2}(4.7\times 10^{-3} \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})(16.24 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2})^{2}-0=0.6 \; \mathrm {J} $$, $$ T_{1}-T_{2}+g(m_{2}-m_{1})=a(m_{1}+m_{2}) $$, $$ T_{2}-T_{1}=\frac{I\alpha }{R}=\frac{Ia}{R^{2}} $$, $$ a=\frac{g(m_{2}-m_{1})}{(m_{1}+m_{2}+I/R^{2)}} $$, $$ a=\frac{g(m_{2}-m_{1})}{(m_{1}+m_{2}+M/{2})} $$, $$ \alpha =\frac{g(m_{2}-m_{1})}{R(m_{1}+m_{2}+M/2)} $$, \(\mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}{:} (\mathrm {a})\), $$ I=\displaystyle \frac{1}{2}MR^{2}=\frac{1}{2}(10 \; \mathrm {k}\mathrm {g})(0.2 \; \mathrm {m})^{2}=0.2 \; \mathrm {kg\, m^2} $$, $$ L=I\omega =(0.2 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})(5 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})=1 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2}/\mathrm {s} $$, $$ \omega =\omega _{0}+\alpha t=(5 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})+(0.5 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2})(3 \; \mathrm {s})=6.5 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s} $$, $$ L=I\omega =(0.2 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})(6.5 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})=1.3 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2}/\mathrm {s} $$, $$\tau =I\alpha =(0.2 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})(0.5 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2})=0.1 \; \mathrm {N\, m}$$, $$ W=\frac{1}{2}I\omega ^{2}-\frac{1}{2}I\omega _{0}^{2}=\frac{1}{2}(0.2\,\mathrm {k}\mathrm {g}\,\mathrm {m}^{2})((6.5 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})^{2}-(5 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})^{2})=1.72 \; \mathrm {J} $$, \(\alpha =(4t)\,\mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2}\), $$ \omega =\int \alpha dt=\int 4tdt=2t^{2}+c $$, $$ \omega =(2t^{2}) \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s} $$, $$ I=\displaystyle \frac{2}{5}MR^{2}+MR^{2}=\frac{7}{5}MR^{2}=\frac{7}{5}(4.7 \; \mathrm {k}\mathrm {g})(0.05 \; \mathrm {m})^{2}=0.016 \; \mathrm {kg\, m^2} $$, $$ L=I\omega =(0.016 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})((2t^{2}) \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})=(0.03t^{2}) \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2}/\mathrm {s} $$, $$ \displaystyle \tau =\frac{dL}{dt}=(0.06t) \mathrm {N\, m} $$, $$ L=I_{z}\omega =(0.014 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})(5 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})=0.07 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2}/\mathrm {s} $$, $$ L=I_{y}\omega =(0.01 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})(5 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})=0.05 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2}/\mathrm {s} $$, $$ L=I_{x}\omega =(4\times 10^{-3} \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})(5 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})=0.02 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2}/\mathrm {s} $$, $$ \triangle L=I(\omega _{f}-\omega _{i}) $$, $$ (0.04 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2}/\mathrm {s})=(0.2 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})(\omega _{f}-(5 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})) $$, \(\omega _{f}=5.2 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}.\), \(\mathrm {r}\mathrm {e}\mathrm {v}/\mathrm {s}{:}\,(\mathrm {a})\), \(I_{f}=3 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})\), $$ \omega _{f}=\frac{I_{i}}{I_{f}}\omega _{i}=\frac{(15 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2}/\mathrm {s})}{(3 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2}/\mathrm {s})}(0.3 \; \mathrm {r}\mathrm {e}\mathrm {v}/\mathrm {s})=1.5 \; \mathrm {r}\mathrm {e}\mathrm {v}/\mathrm {s} $$, $$ \omega _{i}=\bigg (0.3 \; \frac{\mathrm {r}\mathrm {e}\mathrm {v}}{\mathrm {s}}\bigg ) \bigg (\frac{2\pi \; \mathrm {r}\mathrm {a}\mathrm {d}}{1 \; \mathrm {r}\mathrm {e}\mathrm {v}}\bigg )=1.9 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s} $$, $$ \omega _{f}=\bigg (1.5 \; \frac{\mathrm {r}\mathrm {e}\mathrm {v}}{\mathrm {s}}\bigg ) \bigg (\frac{2\pi \; \mathrm {r}\mathrm {a}\mathrm {d}}{1 \; \mathrm {r}\mathrm {e}\mathrm {v}}\bigg )=9.4 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s} $$, $$ K_{i}=\frac{1}{2}I_{i}\omega _{i}^{2}=\frac{1}{2}(15 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})(1.9 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})^{2}=27 \; \mathrm {J} $$, $$ K_{f}=\frac{1}{2}I_{f}\omega _{f}^{2}=\frac{1}{2}(3 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})(9.4 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})^{2}=132.5 \; \mathrm {J} $$, $$ I_{1}\omega _{1}=(I_{1}+I_{2})\omega $$, $$ \omega =\frac{I_{1}\omega _{1}}{(I_{1}+I_{2})}=\frac{(0.1 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})(3 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})}{(0.15 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})}=2 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s} $$, $$ K_{i}=\frac{1}{2}I_{1}\omega _{1}^{2}=\frac{1}{2}(0.1 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})(3 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})^{2}=0.45 \; \mathrm {J} $$, $$ K_{f}=\frac{1}{2}(I_{1}+I_{2})\omega ^{2}=\frac{1}{2}(0.15 \; \mathrm {k}\mathrm {g}\,\mathrm {m}^{2})(2 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s})^{2}=0.3 \; \mathrm {J} $$, \(\alpha =2t \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2}\), https://doi.org/10.1007/978-3-030-15195-9_7, Advances in Science, Technology & Innovation.