Question 769: Centre of Mass and Rotational Motion

Question

Two solid cylinders P and Q of same mass and same radius start rolling down a fixed inclined
plane from the same height at the same time. Cylinder P has most of its mass concentrated near
its surface, while Q has most of its mass concentrated near the axis. Which statement(s) is(are)
correct?

Accepted Answer

Since Cylinder P has most of its mass concentrated near its surface, its moment of inertia (about the cylinder axis) is greater than that of Cylinder Q, i.e., $ I_P > I_Q $. The forces acting on the cylinder include its weight $ mg $, the normal reaction $ N $, and the frictional force $ f $.
In the case of rolling without slipping, we have :
$ v = \omega r $
$ a = \alpha r $
The torque about the center of mass is related to $\alpha$ by :
$ 	au = rf = I\alpha \quad (1) $
The force along the plane is related to $a$ by :
$ mg \sin 	heta - f = ma \quad (2) $
By solving equations (1) and (2), we get :
$ a = \frac{g \sin 	heta}{1 + \frac{I}{mr^2}} \quad (3) $
From equation (3), it follows that $ a_P  I_Q $). Therefore, Cylinder P reaches the ground later, has a lower velocity, lower angular velocity ($\omega = \frac{v}{r}$), and lower translational kinetic energy.
For verification, the conservation of energy can be used :
$ mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 $
This yields :
$ \frac{1}{2}mv_P^2  t_Q $

Answer

Both cylinders P and Q reach the ground at the same time.

Answer

Cylinders P has larger linear acceleration than cylinder Q.

Answer

Both cylinders reach the ground with same translational kinetic energy.

Answer

Cylinder Q reaches the ground with larger angular speed.

Centre of Mass and Rotational Motion

Two solid cylinders P and Q of same mass and same radius start rolling down a fixed inclined plane from the same height at the same time. Cylinder P has most of its mass concentrated near its surface, while Q has most of its mass concentrated near the axis. Which statement(s) is(are) correct?

Choose the correct answer: