Question

A uniform, thin cylindrical shell and solid cylinder roll horizontal without slipping. The speed of the cylindrical shell is v. The solid cylinder and the hollow cylinder encounter an incline that they climb without slipping. If the maximum height they reach is the same, find the initial speed v’ of the solid cylinder.
A. $\sqrt{\dfrac{4}{3}}v$
B. $\sqrt{\dfrac{3}{4}}v$
C. $\sqrt{\dfrac{3}{5}}v$
D. $\sqrt{\dfrac{5}{3}}v$

Answer 1

When both the cylinders climb the incline, their kinetic energy is converted into potential energy and when they reach the maximum their kinetic energy becomes zero. Find the initial kinetic energies of both the cylinders and equate to the potential energy at the maximum height.

Formula used:
$K=\dfrac{1}{2}m{{v}^{2}}+\dfrac{1}{2}I{{\omega }^{2}}$
$v=\omega R$
$U=mgh$
Where v is the speed of the body, I is the moment of inertia about the axis of rotation passing through its centre of mass and $\omega$ is the angular velocity of the body about the same axis.

Complete step by step answer:
Let us assume that the cylindrical shell and the solid cylinder have the same radius R and same mass m. It is given that both the cylinders are rolling without slipping on the horizontal plane. The speed of the cylindrical shell is v and the speed of the solid cylinder is v’.
The kinetic energy of a body when it is rolling is equal to $K=\dfrac{1}{2}m{{v}^{2}}+\dfrac{1}{2}I{{\omega }^{2}}$ … (i),
Where v is the speed of the body, I is the moment of inertia about the axis of rotation passing through its centre of mass and $\omega$ is the angular velocity of the body about the same axis.
And if the circular body is rolling without slipping then $v=\omega R$.
$\Rightarrow \omega =\dfrac{v}{R}$.
Substitute this value in (i).
$\Rightarrow K=\dfrac{1}{2}m{{v}^{2}}+\dfrac{1}{2}I{{\left( \dfrac{v}{R} \right)}^{2}}$ $\Rightarrow K=\dfrac{1}{2}m{{v}^{2}}+\dfrac{1}{2}I{{\left( \dfrac{v}{R} \right)}^{2}}$.
In this case, both the cylinders will be rolling about their standard axes. The moment of inertia of the cylindrical shell about its axis is equal to $I=m{{R}^{2}}$.The moment of inertia of the solid cylindrical about its axis is equal to $I=\dfrac{m{{R}^{2}}}{2}$.
Therefore, the kinetic energy of the cylindrical shell is, ${{K}_{1}}=\dfrac{1}{2}m{{v}^{2}}+\dfrac{1}{2}m{{R}^{2}}{{\left( \dfrac{v}{R} \right)}^{2}}=2\left( \dfrac{1}{2}m{{v}^{2}} \right)=m{{v}^{2}}$.
And the kinetic energy of the solid cylinder is ${{K}_{2}}=\dfrac{1}{2}mv{{'}^{2}}+\dfrac{1}{2}\left( \dfrac{m{{R}^{2}}}{2} \right){{\left( \dfrac{v'}{R} \right)}^{2}}=\dfrac{1}{2}mv{{'}^{2}}+\dfrac{1}{4}mv{{'}^{2}}=\dfrac{3}{4}mv{{'}^{2}}$.
Then they both move up an inclined plane without slipping and reach at equal heights. When both the cylinders climb the incline, their kinetic energy is converted into potential energy and when they reach the maximum their kinetic energy becomes zero.
If the maximum height reach is h, then the potential energy at that height will be $U=mgh$,
where m is the mass of the body and g is acceleration due to gravity. This means that the energy of the body is conserved. i.e. The initial kinetic energy is equal to the potential energy at the maximum height.Therefore,
${{K}_{1}}=U=mgh$ and ${{K}_{2}}=U=mgh$
$\Rightarrow {{K}_{1}}={{K}_{2}}$
$\Rightarrow m{{v}^{2}}=\dfrac{3}{4}mv{{'}^{2}}$
$\Rightarrow v{{'}^{2}}=\dfrac{4}{3}{{v}^{2}}$
$\therefore v'=\sqrt{\dfrac{4}{3}}v$.
This means that the initial speed of the solid cylinder is $\sqrt{\dfrac{4}{3}}v$.

Hence, the correct option is A.

Note: When the cylinder is rolling, the net velocity of the bottommost point (i.e. the point that is in contact with the ground) is zero. And this is due to the friction between the cylinder and the ground. However, some students may think that the friction will do some negative work on the body which is not true in the case of rolling. The work done by the friction is zero because the net displacement of the point of contact is zero and if the displacement of the point at which the force acts is zero then the work done is zero.Hence, the energy is conserved.