Question

A horizontal plat form in the shape of a circular disk rotates in a horizontal plane about a frictionless vertical axle. The plat form has Mass m=100kg and radius r=2m. A students whose mass m=60kg walks slowly on the plat form towards the center. If the angular velocity of the system is 2rad/s when the students at the edge. Then calculate the angular velocity when the students has reached at point 0.5m from the centre

Answer 1

The conservation of angular momentum can be used to solve this problem:

L_i = L_f

where
L_i = initial angular momentum of the system
L_f = final angular momentum of the system

Initially, the platform is rotating with an angular velocity of 2 rad/s and no external torques are acting on the system. Therefore,

L_i = I * w_i

where
I = moment of inertia of the platform
w_i = initial angular velocity of the platform

The moment of inertia of a circular disk is I = (1/2) * m * r^2. Substituting the given values,

L_i = (1/2) * 100 kg * (2 m)^2 * 2 rad/s
= 400 kg m^2/s

When the student reaches a point 0.5 m from the center, the moment of inertia of the system changes because of the redistribution of mass. However, the angular momentum of the system remains conserved. Therefore,

L_i = L_f

I_i * w_i = I_f * w_f

where
I_i = initial moment of inertia of the system (with the student at the edge)
w_i = initial angular velocity of the system
I_f = final moment of inertia of the system (with the student at 0.5 m from the center)
w_f = final angular velocity of the system (to be determined)

The moment of inertia of the system with the student at the edge is I_i = (1/2) * (100+60) kg * (2 m)^2 = 320 kg m^2. The moment of inertia of the system with the student at 0.5 m from the center can be calculated using the parallel axis theorem as follows:

I_f = I_c + m * d^2

where
I_c = moment of inertia of the disk about its own center of mass (i.e., without the student)
m = mass of the student
d = distance between the center of the disk and the student's location

The moment of inertia of the disk about its own center of mass is I_c = (1/2) * 100 kg * (2 m)^2 = 200 kg m^2. Substituting the given values,

I_f = 200 kg m^2 + 60 kg * (0.5 m)^2
= 215 kg m^2

Substituting all the known values into the conservation of angular momentum equation and solving for w_f,

I_i * w_i = I_f * w_f

320 kg m^2 * 2 rad/s = 215 kg m^2 * w_f

w_f = 2.37 rad/s (rounded to two significant figures)

Therefore, the angular velocity of the system when the student has reached a point 0.5 m from the center is 2.37 rad/s.