Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 12 - Rotation of a Rigid Body - Exercises and Problems - Page 350: 56

Answer

See the detailed answer below.

Work Step by Step

a) We know that the moment of inertia is given by $$I=\int x^2 dm$$ And in our case, we are dealing with disks and radiuses, so $$I=\int r^2 dm\tag 1$$ We know that the density of the disk is uniform, so $$\rho=\dfrac{M}{V}=\dfrac{dm}{dV}$$ Hence, $$ \dfrac{M}{ \color{red}{\bf\not}tA}=\dfrac{dm}{ \color{red}{\bf\not}tdA }$$ where $t$ is the thickness of the disk which is constant and $A$ is the area of the disk. Thus, $$dm=\dfrac{MdA}{ A}$$ Recall that the area of a disk's surface (the area of a circle) is given by $A=\pi r^2$. So that the area of the disk that is empty from the middle as our disk is given by $A=\pi (R^2-r^2)$. Thus, $$dm=\dfrac{MdA}{ \pi (R^2-r^2)} $$ where $dA$ is given by $2\pi r\times dr$ $$dm=\dfrac{M (2 \color{red}{\bf\not}\pi r)dr}{ \color{red}{\bf\not}\pi (R^2-r^2)} $$ Plugging into (1); $$I=\int_r^R r^2 \dfrac{2M rdr}{ (R^2-r^2)} $$ $$I= \dfrac{M2 }{ (R^2-r^2)}\int_r^R r^3dr $$ Noting that $A_{disk}=A$ is constant and is given by $ \pi (R^2-r^2)$ that's why we pull it out of the integral. $$I= \dfrac{2M }{ 4 (R^2-r^2)}\bigg|_r^R r^4 $$ $$I= \dfrac{M }{ 2(R^2-r^2)}(R^4-r^4)$$ $$I=\dfrac{M }{ 2 (R^2-r^2)}(R^2-r^2)(R^2+r^2)$$ $$\boxed{I=\dfrac{M(R^2+r^2) }{ 2}}\tag 2$$ _____________________________________________ b) When $r=0$, $$I=\dfrac{M(R^2+0^2) }{ 2}=\boxed{\frac{1}{2}MR^2}$$ which is the same result of the moment of inertia of a sold disk as Table 12.2. And when $r=R$, $$I=\dfrac{M(R^2+R^2) }{ 2}=\boxed{ MR^2}$$ which is the same result of the moment of inertia of a cylindrical hoop, about center as Table 12.2. _____________________________________________ c) We assume that the disk is rolling down the incline without friction which means that the energy is conserved. Thus, $$E_i=E_f$$ $$K_i+K_{i,rot}+U_{gi}=K_f+K_{f,rot}+U_{gf}$$ We choose at the bottom of the incline to be our origin at which $y=0$. Thus, the final potential energy is zero. $$K_i+K_{i,rot}+U_{gi}=K_f+K_{f,rot}+0$$ We know that the disk starts rolling from rest, so that the initial kinetic energy is zero. $$0+0+U_{gi}=K_f+K_{f,rot}$$ $$ Mgh=\frac{1}{2}Mv^2+\frac{1}{2}I\omega^2$$ Plugging $I$ from (2); $$ \color{red}{\bf\not}Mgh=\frac{1}{2} \color{red}{\bf\not}Mv^2+\frac{1}{2}\left[\dfrac{ \color{red}{\bf\not}M(R^2+r^2) }{ 2}\right]\omega^2$$ $$ gh=\frac{1}{2} v^2+ \dfrac{ R^2+r^2 }{ 4} \omega^2$$ We can find the initial height $h$ from the right triangle below, $\sin\theta=\dfrac{h}{d}$. Hence, $h=d\sin\theta$; $$ gd\sin\theta=\frac{1}{2} v^2+ \dfrac{ R^2+r^2 }{ 4} \omega^2$$ Recall that $v=\omega R$, so $\omega=v/R$. Thus, $$ gd\sin\theta=\frac{1}{2} v^2+ \dfrac{ R^2+r^2 }{ 4} \left(\dfrac{v}{R}\right)^2$$ $$ gd\sin\theta=\frac{1}{2} v^2+ \dfrac{ R^2+r^2 }{ 4} \left(\dfrac{v^2}{R^2}\right)$$ $$ gd\sin\theta=v^2\left[\frac{1}{2} + \dfrac{R^2+r^2 }{4R^2} \right]$$ Therfore, the velocity of the disk at the bottom of the incline is given by $$v =\sqrt{\dfrac{gd\sin\theta}{\frac{1}{2} + \dfrac{R^2+r^2 }{4R^2} }}$$ Plugging the known; $$v =\sqrt{\dfrac{9.8\times 0.5\sin20^\circ }{\frac{1}{2} + \dfrac{0.04^2+0.03^2 }{4\times 0.04^2} }}$$ $$v=\color{red}{\bf 1.372 }\;\rm m/s\tag 3$$ If the object was sliding rather than rolling, we can use the same conservation of fnenegy principle with no rotational kinetic energy. Thus, $$ \color{red}{\bf\not}Mgh=\frac{1}{2} \color{red}{\bf\not}M{v^2}'$$ $$v'=\sqrt{2gh}=\sqrt{2gd\sin\theta}$$ Plugging the known; $$v'= \sqrt{2\times 9.8\times 0.5 \sin20^\circ}=\color{red}{\bf 1.831}\;\rm m/s$$ Therefore, $$\dfrac{v}{v'}=\dfrac{1.372 }{1.831}=0.749\approx 0.75$$ $$v={\bf 0.75}v'$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.