A student sits on a freely rotating stool holding two weights, each of mass 2.98

Question

A student sits on a freely rotating stool holding two weights, each of mass 2.98 kg. When his arms are extended horizontally, the weights are 1.06 m from the axis of rotation and he rotates with an angular speed of 0.749 rad/s. The moment of inertia of the student plus stool is 2.98 kg·m2 and is assumed to be constant. The student pulls the weights inward horizontally to a position 0.309 m from the rotation axis.

(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 weights inward.
J (before)
J (after)

Explanation / Answer

a.)

conserve momentum

I1W1 = I2W2 ----1

I is moment of inertia

W is angular velocity

initially moment of interia of the system

I1 = The moment of inertia of the student plus stool + weights

I1 = 2.98 + 2*MR^2 = 2.98 + 2* 2.98kg*1.06^2 = 9.7676 kgm^2

I2 = The moment of inertia of the student plus stool + weights

I2 = 2.98 + 2*MR^2 = 2.98 + 2* 2.98kg*0.309^2 = 3.54 kgm^2

put the value of I 1 and I2 in equn 1

I1W1= I2W2

9.7676 * 0.749 = 3.54* W2

W2 = 2.066 rad/sec

b.)

K.E = 0.5 I2W2^2 = 0.5*3.54*2.066^2 = 7.554 J after

K.E = 0.5I1W1^2 = 2.739 J before

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