Question

A student sits on a freely rotating stool holding two weights, each of which has a mass of 4.81kg. When his arms are extended horizontally, the weights are 1.030m from the axis of rotation and he rotates with an angular speed of 0.803rad/s. The moment of inertia of the student plus stool is 4.46kgm^2 and is assumed to be constant. The student pulls the weights inward horizontally to a position 0.340m from the rotation axis. Find the new angular speed of the student.Calculate the kinetic energy of the system before the student pulls his arms inwards.Calculate the final kinetic energy of the system.

Answer 1

The initial moment of inertia of the weight, Iw = 2 * m * R²
Where m is the mass of each weight and R is the distance from the axis of rotation.
Iw = 2 * 4.81 * 1.03²
= 10.21 kg.m²
Moment of inertia of the student+stool, Is = 4.46 kg.m²
Total Initial moment of inertia, Ii = Iw + Is
Ii = 10.21 + 4.46
= 14.67 kg.m²

Final moment of inertia of the weight, Iw' = 2 * m * r²
Where r is the new distance from the axis of rotation.
Iw = 2 * 4.81 * 0.34²
= 1.112 kg.m²
Moment of inertia of the student+stool, Is = 4.46 kg.m²
Total final moment of inertia, If = Iw' + Is
If = 1.112 + 4.46
= 5.572 kg.m²

a)
Using conservation of angular momentum,
Initial angular momentum = Final angular momentum
Ii * i = If * f
Where i and f are the initial and final angular momentum.
f = (Ii * i) / If
= (14.67 * 0.803) / 5.572
= 2.11 rad/s

b)
Initial kinetic energy, KEi = 1/2 * Ii * i²
= 0.5 * 14.67 * 0.803²
= 4.728 J

c)
Final kinetic energy, KEf = 1/2 * If * f²
= 0.5 * 5.572 * 2.11²
= 12.45 J