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A student on a piano stool rotates freely with an angular speed of 2.85 rev/s . The student holds a 1.50 kg mass in each outstretched arm, 0.789 m from the axis of rotation. The combined moment of inertia of the student and the stool, ignoring the two masses, is 5.53 kg⋅m2 , a value that remains constant.As the student pulls his arms inward, his angular speed increases to 3.60 rev/s . How far are the masses from the axis of rotation at this time, considering the masses to be points

User Shl
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1 Answer

2 votes

Answer:r'=0.327 m

Explanation:

Given


N=2.85 rev/s

angular velocity
\omega =2\pi N=17.90 rad/s

mass of objects
m=1.5 kg

distance of objects from stool
r_1=0.789 m

Combined moment of inertia of stool and student
=5.53 kg.m^2

Now student pull off his hands so as to increase its speed to 3.60 rev/s


\omega _2=2\pi N_2


\omega _2=2\pi 3.6=22.62 rad/s

Initial moment of inertia of two masses
I_0=2mr_^2


I_0=2* 1.5* (0.789)^2=1.867

After Pulling off hands so that r' is the distance of masses from stool


I_0'=2* 1.5* (r')^2

Conserving angular momentum


I_1\omega =I_2\omega _2


(5.53+1.867)\cdot 17.90=(5.53+I_o')\cdot 22.62


I_0'=1.397* 0.791


I_0'=5.851


5.53+2* 1.5* (r')^2=5.851


2* 1.5* (r')^2=0.321


r'^2=0.107009


r'=0.327 m

User Cvsguimaraes
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