On an old-fashioned rotating piano stool, a woman sits holding a pair of dumbbel

Question

On an old-fashioned rotating piano stool, a woman sits holding a pair of dumbbells at a distance of 0.630 m from the axis of rotation of the stool. She is given an angular velocity of 2.75 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 5.00 kgm2 and may be considered constant. Each dumbbell has a mass of 5.20 kg and may be considered a point mass. Neglect friction. A) What is the initial angular momentum of the system? B) What is the initial angular momentum of the system? C) Compute the kinetic energy of the system before the dumbbells are pulled in. D) Compute the kinetic energy of the system after the dumbbells are pulled in.

Explanation / Answer

part A and part B

initial angular momentum of the system L is

L=(Iwoman+Idumbells)

L = Itotal*1

where moment of inertia I=(2)mr2

The woman's moment of inertia Iwoman = 5.0 kg m^2
The woman's moment of inertia Idumbells =(2)mr2 = 2*5.20*r2 = 10.4*0.6302 = 4.12 kgm^2
Itotal = 5.0kg.m2+4.12 kg.m2 = 9.12 kgm^2
L = Itotal*1 = 9.12*2.75 = 25.10 kgm^2/sec

part C

kinetic energy of the system before the dumbbells are pulled

KE1 = 1/2*Itotal*^2 = 0.5*9.12kg.m2 * 2.752

KE1 = 34.38 joule

part D

kinetic energy of the system after the dumbbells are pulled in KE2 = 1/2*Itotal2*22

The woman's moment of inertia Iwoman = 5.0 kg m^2

The woman's moment of inertia Idumbells Id2 = 2*5.2*d2^2 = 10.4*0.165^2 = 0.283 kgm^2
Itotal2 = 5+0.283 = 5.283 kgm^2
L = 25.10 is conserved
2 = L/It2 = 25.10/5.283 = 4.75 rad/sec

KE2 = 1/2*Itotal2*22 = 0.5*5.283 kgm^2 * (4.75 rad/sec)2

KE2 = 59.62JOULES

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