A student sits on a rotating stool holding two 2.1-kg objects. When his arms are
ID: 1441812 • Letter: A
Question
A student sits on a rotating stool holding two 2.1-kg objects. When his arms are extended horizontally, the objects are 1.0 m from the axis of rotation and he rotates with an angular speed of 0.75 rad/s. The moment of inertia of the student plus stool is 3.0 kg · m2 and is assumed to be constant. The student then pulls in the objects horizontally to 0.38 m from the rotation axis.
(a) Find the new angular speed of the student.
________ rad/s
(b) Find the kinetic energy of the student before and after the objects are pulled in.
before _______ J
after ________ J
Explanation / Answer
A: Moment of Inertia at start is: I= 3 + (2*2.1*1^2) = 7.2 kg-m^2
So,Angular momentum (L) = 7.2*0.75 = 5.4 kg m^2 / s
Now,
Moment of Inertia at end is:I' = 3 + (2*2.1*0.38^2) = 3.61 kg-m^2
Hense, From conservation of angular momentum:
New angular velocity = 5.4 / 3.61 = 1.496 rad/s
B:
original KE = 1/2 . MI . w^2 = 0.5 * 7.2 * 0.75^2 = 2.025 J
And,
Final KE = 0.5 * 3.61 * 1.496^2 = 4.04 J
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