A student sits on a freely rotating stool holding two dumbbells, each of mass 3.

Question

A student sits on a freely rotating stool holding two dumbbells, each of mass 3.04 kg (see figure below). When his arms are extended horizontally (Figure a), the dumbbells are 0.97 m from the axis of rotation and the student rotates with an angular speed of 0.747 rad/s. The moment of inertia of the student plus stool is 2.65 kg · m2 and is assumed to be constant. The student pulls the dumbbells inward horizontally to a position 0.296 m from the rotation axis (Figure b).

(a) Find the new angular speed of the student.
rad/s

(b) Find the kinetic energy of the rotating system before and after he pulls the dumbbells inward.

Explanation / Answer

We know that,Angular momentum is conserved.So, using this theory:-
I.w = ( 2.65 + (2 * 3.04 * 0.97^2))*0.747 = 6.25 kgm^2/s
I: moment of inertia & w: Angular velocity
Now,
I(new)*w(new) = 6.25 = (2.65 + (2*3.04*0.296^2))* w(new) = 3.183w(new)
So,
w(new) = 1.96 rad/s

b) Kinetic energy of the rotating system will be 1/2 * I * w^2 each time.
So,
Kbefore= 1/2 * I * w^2= 2.334 J
Kafter = 1/2 * I(new) * w(new)^2= 6.125 J

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