Question

A horizontal platform in the shape of a circular disk rotates on a frictionless bearing about a vertical axle through the center of the disk. The platform has a mass of 127.0 kg, a radius of 2.00 m, and a rotational inertia of 5.08×102 kgm2 about the axis of rotation. A student of mass 66.0 kg walks slowly from the rim of the platform toward the center. If the angular speed of the system is 1.31 rad/s when the student starts at the rim, what is the angular speed when she is 0.740 m from the center?

Answer 1

Use the principle of conservation of angular momentum to solve this problem.

When student is on the rim then the angular momentum (Id + Is)w.

Here, we have -

Id = rotational inertia of the disc = 5.08 x 10^2 kg*m^2 = 508.0 kg*m^2

Is = inertia of student about axis = m x r^2

= 66 x 2^2 = 264.0 kg*m^2

So, the initial angular momentum of the system is given as -

Pi = (508 + 264)*1.31 kg*m^2/s -------------------------------------------------(i)

When the student at a distance of 0.740 m,

Total inertia = 508 + 66 x 0.74^2

= 508 + 36.14 = 544.14 kg*m^2

Hence the angular momentum with new angular speed w1 is given as -

Pf = 544.14 * w1 ----------------------------------------(ii)

Apply conservation of angular momentum -

Pi = Pf

=> (508 + 264)*1.31 = 544.14 * w1

=> w1 = (772 x 1.31) / 544.14 = 1.86 rad/s

Hence, the final angular speed = 1.86 rad/s (Answer)