Question

A disc of radius $2m$ and mass $100kg$ rolls on a horizontal floor. Its centre of mass has a speed of $20cm/s$. How much work is needed to stop it?
(A) $3{\text{J}}$
(B) $30{\text{kJ}}$
(C) $2{\text{J}}$
(D) ${\text{1J}}$

Answer 1

Hint: To solve this question, we need to find out the total kinetic energy of the disc. As the disc is rolling as well as translating, so the total kinetic energy will be the sum of rotational and translational kinetic energy. Then using the work energy theorem we can find out the required work done to stop the disc.
Formula used: The formulae used for solving this question are given by
$I = \dfrac{{M{R^2}}}{2}$, here $I$ is the moment of inertia of a disc of mass $M$ and radius $R$ about the axis passing through its centre and perpendicular to its plane.
${K_R} = \dfrac{1}{2}I{\omega ^2}$, here ${K_R}$ is the rotational kinetic energy, $I$ is the moment of inertia, and $\omega $ is the angular velocity.
\[{K_T} = \dfrac{1}{2}m{v^2}\], here \[{K_T}\] is the translational kinetic energy of a body of mass $m$ moving with a velocity of $v$.

Complete step-by-step solution:
Let $W$ be the work done required to stop the disc. We know from the work energy theorem that the work done is equal to the change in kinetic energy of a body. Let ${K_i}$ and ${K_f}$ be the initial and the final kinetic energies of the disc. So we have
$W = {K_f} - {K_i}$......... (1)
Now, since the disc has been stopped, its final kinetic energy must be zero. So we have
${K_f} = 0$..................(2)
We know that the rotational kinetic energy is given by
${K_R} = \dfrac{1}{2}I{\omega ^2}$....................................(3)
Since a disc rolls about the axis perpendicular to its plane and passing through its centre, so its moment of inertia is given by
$I = \dfrac{{M{R^2}}}{2}$
Substituting this in (3) we have
\[{K_R} = \dfrac{{M{R^2}{\omega ^2}}}{4}\]
\[{K_R} = \dfrac{{M{{\left( {\omega R} \right)}^2}}}{4}\] .........(4)
Assuming that the disc does not slip on the horizontal floor, we can have
$v = \omega R$
Substituting this in (4) we get
\[{K_R} = \dfrac{{M{v^2}}}{4}\] ......... (5)
Now, the translational kinetic energy is given by
\[{K_T} = \dfrac{1}{2}M{v^2}\] ......... (6)
As the disc is rolling as well as translating, so the total initial kinetic energy is given by
${K_i} = {K_R} + {K_T}$
Putting (5) and (6) in above equation, we have
${K_i} = \dfrac{{M{v^2}}}{4} + \dfrac{1}{2}M{v^2}$
$ \Rightarrow {K_i} = \dfrac{3}{4}M{v^2}$......... .................. (7)
Putting (2) and (7) in (1) we get’
$W = 0 - \dfrac{3}{4}M{v^2}$
$ \Rightarrow W = - \dfrac{3}{4}M{v^2}$......... ..................(8)
According to the question, the velocity of the centre of mass of the disc is
$v = 20cm/s$
$ \Rightarrow v = 0.2m/s$.........(9)
Also, the mass of the disc is given as
$M = 100kg$ .................. (10)
Putting (9) and (10) in (8) we get
\[W = - \dfrac{3}{4} \times 100 \times {0.2^2}\]
On solving we finally get
$W = - 3{\text{J}}$
Thus, the magnitude of the work done to stop the disc is equal to $3{\text{J}}$.

Hence, the correct answer is option A.

Note: Do not forget that along with rolling, the disc is also translating forward. So the kinetic energy will have two components; rotational and kinetic. The negative sign of the work done shows that the force applied to stop the disc is opposite to the displacement of the disc.