A student sits on a freely rotating stool holding two dumbbells, each of mass 2.

Question

A student sits on a freely rotating stool holding two dumbbells, each of mass 2.95 kg (see figure below). When his arms are extended horizontally (Figure a), the dumbbells are 0.92 m from the axis of rotation and the student rotates with an angular speed of 0.758 rad/s. The moment of inertia of the student plus stool is 2.77 kg · m2 and is assumed to be constant. The student pulls the dumbbells inward horizontally to a position 0.292 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. Kbefore = J Kafter = J

Explanation / Answer

a.) When in stretched position, the total moment of inertia I1 = moment of Inertia of student and stool + MOI of dumbells

= 2.77 + mR2 + mR2 = 2.77 + 2.95x0.922 + 2.95 x 0.922 = 7.76376 kg m2

When in second position, the total moment of inertia I2 = moment of Inertia of student and stool + MOI of dumbells

= 2.77 + mr2 + mr2 = 2.77 + 2.95x0.2922 + 2.95 x 0.2922 = 3.2730576 kg m2

According to the law of conservation of angular momentum, the total angular momentum before and after should be equal.

So, I11 = I22

7.76376 x 0.758 = 3.2730576 x 2

2 = 1.797991603 rad/ sec

b.) Rotational kinetic energy is given by 0.5 I2

So, KEbefore = 0.5 I112 = 0.5 x 7.76376 x  0.7582 = 2.2303885 J

KEafter =  0.5 I222 = 0.5 x  3.2730576 x 1.7979916032 = 5.290527435 J

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