Question

The following bodies are made roll up (without slipping)the same inclined plane: (i) a ring of radius R, (ii) a solid cylinder of radius $\dfrac{R}{2}$ and (iii) a solid sphere of radius $\dfrac{R}{4}$. If in each case, the speed of the centre of mass at the bottom of the incline is same, the ratio of the maximum heights they climb is:
a)4:2:3
b) 20:15: 14
c) 10:15:7
d) 2:3:4

Answer 1

In the above question it is given to us that different bodies roll along a surface and all of them have the same speed about their respective center of mass when they finally roll up on the incline to a particular height. When the body rolls along a surface it has both, the translational kinetic energy and the rotational kinetic energy. Hence the body will roll such that all the kinetic energy of the bodies get converted to the potential energy at a particular height. To account for both the translational and rotational kinetic energy we have to consider the energy of the body along it center of rotation that makes contact with the surface. Therefore we will equate this energy to the potential energy to obtain the respective height of the bodies.
Formula used:
$U=mgh$
$K.E=\dfrac{1}{2}{{I}_{COR}}{{\omega }^{2}}$
${{I}_{COR}}={{I}_{CM}}+M{{d}^{2}}$
$v=r\omega$

Complete answer:
Let us say a body of mass m rotates and moves up on the incline to a height h. The potential energy(U) of the body is given by $U=mgh$. This potential energy is nothing but the kinetic energy of translational plus the rotational motion of the body. To account for both these energies we have to consider the energy about the center of rotation. If the body has radius R, ${{I}_{COR}}$ is the moment of inertia of the body along its point of rotation i.e. point of contact of the body with the surface and $\omega$ is the angular velocity of the body then its total kinetic energy is given by,
$K.E=\dfrac{1}{2}{{I}_{COR}}{{\omega }^{2}}$
Let us say the velocity of the body along the centre of mass is ‘v’. Hence the above equation becomes,
$\begin{aligned} & K.E=\dfrac{1}{2}{{I}_{COR}}{{\omega }^{2}}\text{, }\because v=R\omega \\\ & \Rightarrow K.E=\dfrac{1}{2}{{I}_{COR}}{{\left( \dfrac{v}{R} \right)}^{2}}=\dfrac{{{I}_{COR}}{{v}^{2}}}{2{{R}^{2}}} \\\ \end{aligned}$
The moment of inertia of a ring of mass M and radius R about its centre of mass is given as, ${{I}_{CM}}=M{{R}^{2}}$ Hence by parallel axis theorem i.e.${{I}_{COR}}={{I}_{CM}}+M{{d}^{2}}$ (d is the distance between the point of contact and center of mass and the) the moment of inertia (${{I}_{COR}}$)of the body along the point of contact is given by,
$\begin{aligned} & {{I}_{COR}}={{I}_{CM}}+M{{d}^{2}} \\\ & \Rightarrow {{I}_{COR}}=M{{R}^{2}}+M{{R}^{2}} \\\ & \Rightarrow {{I}_{COR}}=2M{{R}^{2}} \\\ \end{aligned}$
Hence the total energy of rotation is equal to,
$\begin{aligned} & K.E=\dfrac{{{I}_{COR}}{{v}^{2}}}{2{{R}^{2}}} \\\ & \Rightarrow K.E=\dfrac{2M{{R}^{2}}{{v}^{2}}}{2{{R}^{2}}}=M{{v}^{2}} \\\ \end{aligned}$
Now since this energy becomes equal to the potential energy of the let us say at height ${{h}_{1}}$ we get the height equal to,
$\begin{aligned} & K.E=U({{h}_{1}}) \\\ & \Rightarrow M{{v}^{2}}=Mg{{h}_{1}} \\\ & \Rightarrow {{h}_{1}}=\dfrac{{{v}^{2}}}{g} \\\ \end{aligned}$
The moment of inertia of a solid cylinder about the centre of mass M and radius R/2 is given by ${{I}_{CM}}=\dfrac{M{{(R/2)}^{2}}}{2}=\dfrac{M{{R}^{2}}}{8}$
Hence it’s moment of inertia (${{I}_{COR}}$)of the body along the point of contact is given by,
$\begin{aligned} & {{I}_{COR}}={{I}_{CM}}+M{{d}^{2}} \\\ & \Rightarrow {{I}_{COR}}=\dfrac{M{{R}^{2}}}{8}+M{{(R/2)}^{2}} \\\ & \Rightarrow {{I}_{COR}}=\dfrac{3M{{R}^{2}}}{8} \\\ \end{aligned}$
Hence the total energy of rotation is equal to,
$\begin{aligned} & K.E=\dfrac{{{I}_{COR}}{{v}^{2}}}{2{{(R/2)}^{2}}} \\\ & \Rightarrow K.E=\dfrac{4}{2}\dfrac{3M{{R}^{2}}{{v}^{2}}}{8{{R}^{2}}}=\dfrac{3M{{v}^{2}}}{4} \\\ \end{aligned}$
Now since this energy becomes equal to the potential energy of the let us say at height ${{h}_{2}}$ we get the height equal to,
$\begin{aligned} & K.E=U({{h}_{2}}) \\\ & \Rightarrow \dfrac{3M{{v}^{2}}}{4}=Mg{{h}_{2}} \\\ & \Rightarrow {{h}_{2}}=\dfrac{3{{v}^{2}}}{4g} \\\ \end{aligned}$
The moment of inertia of a solid sphere about the centre of mass M and radius R/4 is given by ${{I}_{CM}}=\dfrac{2M{{(R/4)}^{2}}}{5}=\dfrac{M{{R}^{2}}}{40}$
Hence it’s moment of inertia (${{I}_{COR}}$)of the body along the point of contact is given by,
$\begin{aligned} & {{I}_{COR}}={{I}_{CM}}+M{{d}^{2}} \\\ & \Rightarrow {{I}_{COR}}=\dfrac{M{{R}^{2}}}{40}+M{{(R/4)}^{2}} \\\ & \Rightarrow {{I}_{COR}}=\dfrac{7M{{R}^{2}}}{80} \\\ \end{aligned}$
Hence the total energy of rotation is equal to,
$\begin{aligned} & K.E=\dfrac{{{I}_{COR}}{{v}^{2}}}{2{{(R/4)}^{2}}} \\\ & \Rightarrow K.E=\dfrac{7M{{R}^{2}}}{80}\dfrac{16{{v}^{2}}}{2{{R}^{2}}}=\dfrac{7M{{v}^{2}}}{10} \\\ \end{aligned}$
Now since this energy becomes equal to the potential energy of the let us say at height ${{h}_{3}}$ we get the height equal to,
$\begin{aligned} & K.E=U({{h}_{3}}) \\\ & \Rightarrow \dfrac{7M{{v}^{2}}}{10}=Mg{{h}_{3}} \\\ & \Rightarrow {{h}_{2}}=\dfrac{7{{v}^{2}}}{10g} \\\ \end{aligned}$
Hence the ratio of ${{h}_{1}}:{{h}_{2}}:{{h}_{3}}$is,
$\begin{aligned} & {{h}_{1}}:{{h}_{2}}:{{h}_{3}}=\dfrac{{{v}^{2}}}{g}:\dfrac{3{{v}^{2}}}{4g}:\dfrac{7{{v}^{2}}}{10g} \\\ & \Rightarrow {{h}_{1}}:{{h}_{2}}:{{h}_{3}}=1:\dfrac{3}{4}:\dfrac{7}{10}=40:30:28 \\\ & \Rightarrow {{h}_{1}}:{{h}_{2}}:{{h}_{3}}=20:15:14 \\\ \end{aligned}$

So, the correct answer is “Option B”.

Note:
In the above question it is given to us that the velocity of the three bodies along their center of mass is the same. Hence we have considered v to be the same for all three. The body also is moving on a frictionless surface. Therefore we can say that the total kinetic energy gets converted entirely to the potential energy of the three bodies.