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

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 of the following statement is correct?
a. Both the cylinders P and Q reach the ground at the same time.
b. Cylinder P has larger linear acceleration than cylinder Q
c. Both cylinders reach the ground with same translation kinetic energy.
d. Cylinder Q reaches the ground wih larger angular speed.

TestHub · Accepted Answer

The moment of inertia of a body is given by $I = \int r^{2} d m$ .

For cylinder P, most of its mass is concentrated near the surface, so the effective radius is larger, making its moment of inertia $I_{P}$ larger.

For cylinder Q, most of its mass is concentrated near the axis, so the effective radius is smaller, making its moment of inertia $I_{Q}$ smaller.

Thus, $I_{P} > I_{Q}$ .

The linear acceleration of a body rolling down an inclined plane of angle $θ$ is given by $a = \frac{g sin θ}{1 + \frac{I}{M R ^{2}}}$ .

Since $I_{P} > I_{Q}$ , we have $a_{P} < a_{Q}$ . Cylinder Q has a larger linear acceleration and reaches the ground first.

By conservation of mechanical energy, the velocity at the bottom of the incline is $v = \frac{2 g h}{1 + \frac{I}{M R ^{2}}}$ .

Since $I_{P} > I_{Q}$ , we get $v_{P} < v_{Q}$ . The translational kinetic energy is $\frac{1}{2} M v^{2}$ , so cylinder Q has a larger translational kinetic energy.

The angular speed at the bottom is $ω = \frac{v}{R}$ .

Since $v_{P} < v_{Q}$ , we get $ω_{P} < ω_{Q}$ . Therefore, cylinder Q reaches the ground with a larger angular speed.

Answer: d

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