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Aliun [14]
3 years ago
11

A student sits on a rotating stool holding two 2.8-kg objects. When his arms are extended horizontally, the objects are 1.0 m fr

om 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 0.34 m from the rotation axis. (a) Find the new angular speed of the student.
Physics
1 answer:
Sindrei [870]3 years ago
5 0

Answer:

Explanation:

Given

mass of objects m is 2.8 kg

objects is r=1 m away from axis of rotation

angular  velocity \omega _1=0.75 rad/s

moment of inertia of stool and student is=3 kg-m^2

New distance of objects r'=0.34 m

conserving Angular momentum

Initial Angular momentum L_1=I\omega _1

I_1=3+2\times 2.8\times 1^2

I_1=3+5.6=8.6 kg-m^2

L_1=8.6\times 0.75=6.45

For Second case

I_2=3+2\times 2.8\times 0.34^2=3.32 kg-m^2

L_2=3.32\times \omega _2

L_1=L_2

6.45=3.32\times \omega _2

\omega _2=1.94 rad/s

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