On an old-fashioned rotating piano stool, a woman sits holding a pair of dumbbel
ID: 1553903 • Letter: O
Question
On an old-fashioned rotating piano stool, a woman sits holding a pair of dumbbells at a distance of 0.585 m from the axis of rotation of the stool. She is given an angular velocity of 2.85 rad/s , after which she pulls the dumbbells in until they are only 0.165 m distant from the axis. The woman's moment of inertia about the axis of rotation is 4.60 kgm2 and may be considered constant. Each dumbbell has a mass of 5.50 kg and may be considered a point mass. Neglect friction.
Part A
What is the initial angular momentum of the system?
Part B
What is the angular velocity of the system after the dumbbells are pulled in toward the axis?
Part C
Compute the kinetic energy of the system before the dumbbells are pulled in.
Part D
Compute the kinetic energy of the system after the dumbbells are pulled in.
Explanation / Answer
The initial moment of inertia of the system will be
Ii = I(woman) + I(dumbell)
Ii = 4.6 + 2 x 5.5 x 0.585^2 = 8.36 kg-m^2
Li = I(i) w
Li = 8.36 x 2.85 = 23.83 kg-m^2/s
Hence, Li = 23.83 kg-m^2/s
B)Moment of inertia after the dumbells are pulled in is:
I(f) = I(woman) + I(dumbell)
I(f) = 4.6 + 2 x 5.5 x 0.165^2 = 4.89 kg-m^2/s
from conservation of angular momentum
L(i) = L(f)
Ii wi = If wf
wf = Ii wi/If = 8.36 x 2.85/4.89 = 4.87 rad/s
Hence, wf = 4.87 rad/s
C)KEi = 1/2 Ii wi^2
KEi = 0.5 x 8.36 x 2.85^2 = 33.95 J
Hence, KEi = 33.95 J
D)KEf = 1/2 If wf^2
KEf = 0.5 x 4.89 x 4.87^2 = 57.99 J
Hence, KE = 57.99 J