Question

A uniform cylinder of radius 10 cm and mass 20 kg is mounted so as to rotate freely about a horizontal axis that is parallel to and 5.0 cm from the central longitudinal axis of the cylinder.(a) What is the rotational inertia of the cylinder about the axis of rotation?(b) If the cylinder is released from rest with its central longitudinal axis at the same height as the axis about which the cylinder rotates, what is the angular speed of the cylinder as it passes through its lowest position?

Answer 1

Answer:

a. 0.15 kg m2

b. 19.8 rad/s

Explanation:

Metric unit conversion:

10 cm = 0.1 m

5 cm = 0.05 m

a. Using parallel axis theorem, the rotational inertia of the cylinder about the axis of rotation is the inertia about the longitudinal axis plus the product of mass and distance from the longitudinal axis to the rotational axis squared

$I = I_L + md^2$

whereas the inertia about the longitudinal axis of the solid cylinder is

$I_L = mr^2/2 = 20*0.1^2/2 = 0.1 kgm^2$

$md^2 = 20*0.05^2 = 0.05 kgm^2$

$I = I_L + md^2 = 0.1 + 0.05 = 0.15 kg m^2$

b. Assume the cylinder does not rotate about its own longitudinal axis, we can treat this as a point mass pendulum. So when it's being released from 0.05m high (release point) to 0m (lowest position), its potential energy is converted to kinetic energy:

$E_p = E_k$

$mgh = mv^2/2$

where h = 0.05 is the vertical distance traveled, v is the cylinder linear velocity at  the lowest position.g = 9.81m/s2 is the gravitational acceleration.

We can divide both sides by m

$gh = v^2/2$

$v^2 = 2gh = 2*9.81*0.05 = 0.981$

$v = \sqrt{0.981} = 0.99 m/s$

The angular speed is linear speed divided by the radius of rotation, which is distance from the cylinder center to the center of rotation d = 0.05 m

$\omega = \frac{v}{d} = \frac{0.99}{0.05} = 19.8 rad/s$