A solid disk of mass m1 = 9.6 kg and radius R = 0.21 m is rotating with a consta

Question

A solid disk of mass m1 = 9.6 kg and radius R = 0.21 m is rotating with a constant angular velocity of = 34 rad/s. A thin rectangular rod with mass m2 = 3.5 kg and length L = 2R = 0.42 m begins at rest above the disk and is dropped on the disk where it begins to spin with the disk.

1)

What is the initial angular momentum of the rod and disk system?

2)

What is the initial rotational energy of the rod and disk system?

3)

What is the final angular velocity of the disk?

4)

What is the final angular momentum of the rod and disk system?

5)

What is the final rotational energy of the rod and disk system?

6)

The rod took t = 5.6 s to accelerate to its final angular speed with the disk. What average torque was exerted on the rod by the disk?

Explanation / Answer

The initial rotational energy is (Erotational) =(1/2)Iw2 =(1/2)*(1/2)mw2R2 =(1/4)*9.1kg*(34rad/s)2(0.21)2 =115.978J

The initial angular momentum is given by L1 =Iw =(1/2)mr2w =0.5*9.1kg*(0.21)2*(34rad/s) =6.822kg.m2.rad/s

The final angular velocity is given by

I1w1+I2w2=(I1+I2)wf

0.5*9.1*(0.21)2*(34) =[(0.5*9.1*(0.21)2+0.5*3.5kg*(0.42)2]wf

wf =0.5*9.1*(0.21)2*(34)/[(0.5*9.1*(0.21)2+0.5*3.5kg*(0.42)2]=6.822/[0.5093]=13.395rad/s

The final angualr momentum is given by

We know that final angualr momentum is equal

L2 =(I1+I2)wf =(0.5093)(13.395) =6.822kg.m2.rad/sec

The final rotational energy is given by

Erotational =(1/2)(I1+I2)wf2 =0.5*0.5093*(13.395)2 =45.690J

The average torque is given by

T =I*a

I =(1/2)m2(r2)2

acceleration is given by a =wf/t =13.395/5.6 =2.392rad/s2

Now substitute in the above formula you get the torque

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