Question

Conservation of Angular Momentum

A female figure skater is spinning on ice. Assume that that the surface is basically frictionless. The skater is wearing weighted bracelets as part of the costume for the performance. These weighted bracelets weight .75 kg each. The skater has a spinning routine in the middle of the performance and initially starts spinning with arms stretched wide such that the weighted bracelets are 1 m from the axis of rotation. She has an initial angular velocity of 2LaTeX: \pi π rad/s with an initial moment of inertia of 2.5 kg * m2. The skater then pulls her arms in towards her chest so that the weighted bracelets are just 10 cm from the axis of rotation as this will allow a faster spin, her moment of inertia drops to 2.1 kg * m2.

Part 1. What is the initial angular velocity of this skater. (answer should be a whole number)

rev/s

Part 2. What is the initial kinetic energy the skater has? (give your answer to the nearest J)

J

Part 3. What is the final angular velocity of this skater. (give your answer to 3 significant digits)

rev/s

Part 4. What is the final kinetic energy the skater has? (give your answer to the nearest J)

J

Answer 1

Since the velocity is given in terms of  2LaTeX: \pi π rad/s which is assumed to be 2π rad/s , and As it is not specified that the Moment of inertia given in above questions are including the weights of bracelets also hence we will assume that the given moment of inertia are inclusive of both weight of skater + weight of bracelets.

Part 1)

the angular speed of skater is 2*pi rad/s

Since the required velocity is in rev/ s hence 1 rev = 2*pi hence velocity = 1 rev/ s

Part 2)

kinetic energy = 0.5 * moment of inertia * angular speed^2 = 0.5IW^2

hence initial I = 2.5 W = 2*pi

hence kinetic energy = 49.3 = 49J

Part 3)

Balancing angular momentum of the skater as there is no external force acting on the skater we get

I1W1 = I2W2

2.5 * 2*pi = 2.1 * W2

hence W2 = 2.4 *pi rad/s = 1.2 rev / seconds

Part 4)

Kinetic energy = 0.5*I*W^2

hence final kinetic energy = 0.5* 2.1*(2.4*pi)^2 = 59.6 J = 60J