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) 3J
b) 30kJ
c) 2J
d) 1J

Answer 1

When the disc rolls on a horizontal floor it possesses both the kinetic energy of translation plus the energy of rotational motion. Let us say there is no energy loss due to the friction on the floor. Hence for the disc to stop rolling as well as translate, an equal amount of energy is required to stop its motion
Formula used:
$K.E=\dfrac{1}{2}m{{v}^{2}}$
$R.E=\dfrac{1}{2}I{{\omega }^{2}}$
$v=\omega r$

Complete step by step answer:
In the question it is given that the disc rolls on a horizontal floor whose centre of mass has a speed of 20cm/s. The body translates as well as rotates and hence it will posses angular velocity as well. If the radius of the body is ‘r’, rotates with a angular velocity $\omega$ , then the velocity ‘v’ of the body along the centre of mass is given by,
$v=\omega r$
For a circular disc with radius ‘r’ and mass ‘m’, the moment of inertia along the axis passing through the center is given by,
$I=\dfrac{1}{2}m{{r}^{2}}$
The energy of rotation of a circular body is equal to $R.E=\dfrac{1}{2}I{{\omega }^{2}}$ and the energy of translation is $K.E=\dfrac{1}{2}m{{v}^{2}}$
Hence when the circular disc is rotating as well as translating the total energy ‘E’ is
$\begin{aligned} & E=K.E+R.E \\\ & E=\dfrac{1}{2}m{{v}^{2}}+\dfrac{1}{2}I{{\omega }^{2}} \\\ \end{aligned}$
In order to stop the rolling as well as the displacement of the above body, an equal amount of energy is required to bring the body to rest. Hence the energy ‘E’ is numerically equal to,
$\begin{aligned} & E=\dfrac{1}{2}m{{v}^{2}}+\dfrac{1}{2}I{{\omega }^{2}} \\\ & \because v=\omega r\text{ }\\!\\!\And\\!\\!\text{ }I=\dfrac{1}{2}m{{r}^{2}} \\\ & E=\dfrac{1}{2}m{{v}^{2}}+\dfrac{1}{2}\left( \dfrac{1}{2}m{{r}^{2}} \right){{\left( \dfrac{v}{r} \right)}^{2}} \\\ & \Rightarrow E=\dfrac{1}{2}m{{v}^{2}}+\dfrac{1}{4}m{{v}^{2}} \\\ & \Rightarrow E=\dfrac{3}{4}m{{v}^{2}},\text{ }m=100kg,\text{ }v=20\times {{10}^{-2}}m/s \\\ & \therefore E=300\times {{10}^{-2}}J=3J \\\ \end{aligned}$

So, the correct answer is “Option a”.

Note: It is to be noted that there is no energy lost due to friction. But we also cannot say that there is no friction or else the body will not translate as there will be no torque due to friction. As a consequence, it will not have any speed along the center of mass or more specifically it will just rotate along the axis passing through the center.