Question

a solid cylinder of inertia 1.2 kgm2 rotates about a vertical frictionless axle with angular velocity 25 radians/second. a second solid cylinder of inertia 2.2 kgm2 is initially not rotating and it drops onto the first cylinder. the two surfaces are rough and they eventually come to the same angular velocity. calculate the magnitude of the final angular velocity. answer in units of rad/s.

Answer 1

By applying the conservation of angular momentum, the final angular velocity of the combined system of two solid cylinders is calculated to be 8.824 rad/s.

Step-by-step explanation:

To solve for the final angular velocity after the second solid cylinder drops onto the first, we apply the law of conservation of angular momentum. The total angular momentum before the second cylinder drops must equal the total angular momentum after they have coupled and are spinning at the same angular velocity.

Let I1 = 1.2 kgm2 and \(\omega1\) = 25 rad/s be the moment of inertia and angular velocity of the first cylinder. Let I2 = 2.2 kgm2 be the moment of inertia of the second cylinder which is initially stationary (\(\omega2\) = 0). After coupling, if \(\omegaf\) is the final angular velocity of the system, the conservation of angular momentum gives us:

I1\(\omega1\) + I2\(\omega2\) = (I1 + I2)\(\omegaf\)

Inserting the given values, we get:

1.2 kgm2 * 25 rad/s + 2.2 kgm2 * 0 rad/s = (1.2 kgm2 + 2.2 kgm2) * \(\omegaf\)

\(\omegaf\) = (1.2 * 25) / (1.2 + 2.2) = 30 / 3.4 = 8.824 rad/s

The final angular velocity of the system is 8.824 rad/s.