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Andre45 [30]
4 years ago
10

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 outstre

tched 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
Mathematics
1 answer:
Vlad1618 [11]4 years ago
7 0

Answer:r'=0.327 m

Step-by-step 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\times 1.5\times (0.789)^2=1.867

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

I_0'=2\times 1.5\times (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\times 0.791

I_0'=5.851

5.53+2\times 1.5\times (r')^2=5.851

2\times 1.5\times (r')^2=0.321

r'^2=0.107009

r'=0.327 m

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