Q#1) A) A student sits on a freely rotating stool holding two dumbbells, each of
ID: 1789892 • Letter: Q
Question
Q#1) A) A student sits on a freely rotating stool holding two dumbbells, each of mass 2.97 kg (see figure below). When his arms are extended horizontally (Figure a), the dumbbells are 1.06 m from the axis of rotation and the student rotates with an angular speed of 0.752 rad/s. The moment of inertia of the student plus stool is 2.59 kg · m2 and is assumed to be constant. The student pulls the dumbbells inward horizontally to a position 0.308 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.
(a) What is the magnitude of the angular momentum of the system relative to the center of mass?
kg · m2/s
(b) What is the angular speed about the center of mass?
130 rad/s
Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. rad/s
Explanation / Answer
Q1)Given,
m = 2.97 kg ; r = 1.06 m ; w = 0.752 rad/s ; I = 2.59 kg-m^2 ; r' = 0.308 m
a)The moment of intertia of the system will be:
Isys = I + 2mr^2
Isys(intial) = 2.59 + 2 x 2.97 x 1.06^2 = 9.26 kg-m^2
I1 = 9.26 kg -m^2
when student pulled the dumbell in
Isys(final) = I + 2 mr'^2
Isys(final) = 2.59 + 2 x 2.97 x 0.308^2 = 3.15 kg-m^2
I2 = 3.15 kg-m^2
from conservation of angular momentum
I1w1 = I2w2
w2 = (I1/I2)w1
w2 = (9.26/3.15)0.752 = 2.211 rad/s
Hence, w2 = 2.211 rad/s
KEi = 1/2 I1 w1^2
KEi = 0.5 x 9.26 x 0.752^2 = 2.618 J
KEf = 1/2 I2 w2^2
KEf = 0.5 x 3.15 x 2.211^2 = 7.699 J
Hence, KE(before) = 2.618 J ; KE(after) = 7.699 J
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