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 0.585m from the axis of rotation of the stool. She is given an angular velocity of 3.35rad/s , after which she pulls the dumbbells in until they are only 0.250m distant from the axis. The woman's moment of inertia about the axis of rotation is 4.60kg?m2and may be considered constant. Each dumbbell has a mass of 5.40kg and may be considered a point mass. Neglect friction.

A) What is the initial angualr momentum of the system?

B)What is the angular velocity of the system after the dumbbells are pulled in toward the axis?

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

A) Initially :
Moment of inertia of woman Iw = 4.60 kg.m^2

moment of inertia of dumbells Id = 2 x 5.40 x 0.585^2 = 3.70 kg.m^2

Ii = Iw + Id = 8.30 kg.m^2

angular velocity initial Li = Ii*wi = 8.30 x 3.35 = 27.80

B) Finally:
Moment of inertia of woman Iw = 4.60 kg.m^2

moment of inertia of dumbells Id = 2 x 5.40 x 0.250^2 = 0.675 kg.m^2

Ii = Iw + Id = 5.275 kg.m^2

Lf = 5.275wf

USing angular momentum conservation,

Li = Lf

27.80 = 5.275wf

wf = 5.27 rad/s

c) K.E.i = Ii *wi^2 /2

          = 8.30 x 3.35^2 /2 = 46.57 J

D) K.E.f = If *wf^2 /2

          = 5.275 x 5.27^2 /2 = 73.18 J

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.