A skater extends her arms horizontally, holding a 5-kg mass in each hand. She is rotating about a vertical axis with an angular velocity of one revolution per second. If she drops her hands to her sides, what will the final angular velocity (in rev/s) be if her moment of inertia remains approximately constant at 5 kg · m2, and the distance of the masses from the axis changes from 1 m to. 1 m?

a. 6

b. 3

c. 9

d. 4

e. 7

The answer is b. 3

Explanation:

We use the conservation of angular momentum to solve this problem. This law states that the angular momentum of a rotational mass will remain constant unless there is an external torque acting on it.

Angular momentum is defined by:

L=I*w

where:

L is the angular momentum in [kg*m^2/sec) I is the mass moment of inertia in [kg*m^2] w is the angular velocity in [rev/sec].

Since angular momentum is conserved, we have:

L1 = I1*w1 and L2 = I2 * w2 I1*w1 = I2*w2

We are looking for the final angular velocity:

w2 = I1*w1/I2

Step 1: Calculate I1

We know that the skater's mass moment of inertia is 5 kg*m^2, and that she is holding two 5kg masses in each extended arm at 1m from her rotational axis. We need to add the mass moment of inertia of the two masses to the skater's mass moment of inertia knowing that the mass moment of inertia of an object rotating about an axis is:

I = m*r^2

Therefore:

I1 = I + m*r^2 + m*r^2 I1 = 5 + 2*5*1^2 I1 = 15 [kg*m^2]

Step 2: Calculate I2

Same calculation as step 1, except the masses are now. 1m away from the rotational axis.

I2 = 5 + 2*5*.1^2 I2 = 5.1 [kg*m^2]

Step 3: Calculate w2

We know all the values to input into our w2 equation:

w2 = 15*1/5.1 w2 = 2.94 [rev/sec]

The option that is closer to our answer is 3, therefore that is the correct answer when we round up.