A solid disk of mass m 1 = 9.3 kg and radius R = 0.22 m is rotating with a const

Question

A solid disk of mass m1 = 9.3 kg and radius R = 0.22 m is rotating with a constant angular velocity of = 32 rad/s. A thin rectangular rod with mass m2 = 3.9 kg and length L = 2R = 0.44 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 = 6.7 s to accelerate to its final angular speed with the disk. What average torque was exerted on the rod by the disk?

Explanation / Answer

1)

wdi = initial angular velocity of disk = 32 rad/s

Id = moment of inertia of disk = m1 R2/2 = (9.3) (0.22)2 /2 = 0.2251 kgm2

wri = initial angular velocity of rod = 0 rad/s

Ir = moment of inertia of rod = m2 L2/12 = (3.9) (0.44)2 /12 = 0.063 kgm2

initial angular momentum of rod and disk system :

initial angular momentum = initial angular momentum of rod + initial angular momentum of disk

Li = Id Wdi + Ir Wri

Li = 0.2251 x 32 + 0.063 x 0 = 7.2032

2)

initial rotational energy of rod and disk system :

initial rotational energy = initial rotational energy of rod + initial rotational energy of disk

RE = (0.5) Id W2di + (0.5)Ir W2ri = (0.5) (0.2251 (32)2 + 0.063 x 02 ) = 115.25

c)

Using conservation of angular momentum

final total angular momentum = Li

( Id + Ir ) Wf = Li

(0.2251 + 0.063) Wf = 7.2032

Wf = 25 rad/s

d)

final total angular momentum = Li = 7.2032

e)

final Rotational energy = (0.5)( Id + Ir ) W2f = (0.5) (0.2251 + 0.063) (25)2 = 90.03

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