A student sits on a freely rotating stool holding two dumbbells, each of mass 2.
ID: 777553 • Letter: A
Question
A student sits on a freely rotating stool holding two dumbbells, each of mass 2.93 kg (see figure below). When his arms are extended horizontally (figure a), the dumbbells are 0.90 m from the axis of rotation and the student rotates with an angular speed of 0.760 rad/s. The moment of inertia of the student plus stool is 2.79 kg · m2 and is assumed to be constant. The student pulls the dumbbells inward horizontally to a position 0.293 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) Angular momentum = I , where I is Inertia and is angular velocity in rad/s
For a rotating mass, I = mR^2
So total Inertia of dumbbells = 2 x 2.93 x 0.9 x 0.9 = 4.75 kgm^2
Momentum of dumbbells = I = 4.75 x 0.760 = 3.61 kgm^2/s
Momentum of student + stool = I = 2.79 x 0.760 = 2.12 kgm^2/s
Total (initial) momentum = 5.73 kgm^2/s
Let new angular speed be (2):
Final momentum = 5.73 = 2.79 (2) + [2 x 2.93 x 0.293 x 0.293 x (2)]
5.73 = 2.79 (2) + 0.503(2)
2 = 1.74 rad/s
New angular speed is 1.74 rad/s
b) Rotational KE= 1/2 I ^2
Initial KE = 0.5 x (4.75 + 2.79) x 0.760 x 0.760
= 2.18 J
Final KE = 0.5 x (0.503 +2.79) x 1.72 x 1.72
= 4.98J
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