Question

A uniform disk of mass m and radius R is projected horizontally with velocity $V_0$ on a rough horizontal floor so that it starts off with a purely sliding motion at t = 0. After $t_0$ seconds, it acquires a purely rolling motion as shown in figure.

• A. Angular momentum of disc at t = 0 about a point on the floor and in the same plane which contains the disc is $m{V}_{0}R$.
• B. Angular momentum of disc at $t=t_0$ about a point on the floor and in the same plane which contains the disc is $m{V}_{0}R$.
• C. Speed of the centre of mass of disc at $t=t_0$ is $2V_0/3$
• D. Speed of the centre of mass of disc at $t=t_0$ is $3V_0/4$

Answer 1

Answer: (A), (B), (C)

The correct options are
A Angular momentum of disc at t = 0 about a point on the floor and in the same plane which contains the disc is $m{V}_{0}R$.
B Angular momentum of disc at $t=t_0$ about a point on the floor and in the same plane which contains the disc is $m{V}_{0}R$.
C Speed of the centre of mass of disc at $t=t_0$ is $2V_0/3$

Initial angular velocity = 0.
Let's take a point on the ground,
as friction force passes through this point, its torque is Zero.
Also torque due to N and Mg cancels each other.
As the net torque is zero about this point therefore angular movementum remains conserved about it.

Initial angular momentum = Final angular momentum

To calculate angular momentum we can use the following equation in inertial frame -
Net Angular momentum = Angular momentun of center of mass + Angular momentum about center of mass.

Initial angular momentum
$L_i=m{V}_{0}R$

Final angular momentum
$L_f=mVR+\frac{m{R}^2}{2}\omega$

Now using,
Initial angular momentum = Final angular momentum
and $V=R\omega$for pure rolling on ground, we obtain:
$V=\frac{2}{3}{V}_0$