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Kamila [148]
3 years ago
8

A certain merry-go-round is accelerated uniformly from rest and attains an angular speed of 1.2 rad/s in the first 18 seconds. I

f the net applied torque is 1200 N· m, what is the moment of inertia of the merry-go-round?
A) This cannot be determined since the radius is not given.
B) 1400 kg*m2
C) 18,000 kg*m2
D) 500 kg*m2
E) 9000 kg*m2
Physics
1 answer:
muminat3 years ago
3 0

Answer:

Moment of inertia will be I=18000kgm^2

So option (c) will be correct answer

Explanation:

We have given initial angular velocity \omega _i=0rad/sec

Final angular velocity \omega _f=1.2rad/sec

Time taken to reach final angular velocity t = 18 sec

According to first equation of motion

\omega _f=\omega _i+\alpha t

1.2=0+\alpha \times 18

\alpha =0.066rad/sec^2

Torque is given in question as 1200 N -m

We know that torque is equal to \tau =I\alpha, here I is moment of inertia and \alpha is angular acceleration

So 1200=I\times 0.0666

I=18000kgm^2

So option (c) will be correct answer

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Answer:

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Explanation:

given data

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the total stretch of the double-length spring will be

solution

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k = mg    ............1

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