A solid disk of mass m 1 = 9.5 kg and radius R = 0.19 m is rotating with a const

Question

A solid disk of mass m1 = 9.5 kg and radius R = 0.19 m is rotating with a constant angular velocity of = 39 rad/s. A thin rectangular rod with mass m2 = 3.3 kg and length L = 2R = 0.38 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?

kg-m2/s

2)

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

J

3)

What is the final angular velocity of the disk?

rad/s

4)

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

kg-m2/s

5)

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

J

6)

The rod took t = 6.5 s to accelerate to its final angular speed with the disk. What average torque was exerted on the rod by the disk?

N-m

Explanation / Answer

1) moment of inertia of disc , I1 = 0.5MR^2 =0.5*9.5*0.19^2 = 0.171475

moment of inertia of rod I2 = 1/12 M(2R)^2 = 1/3 * 3.3*0.19^2 = 0.03971

initial angular momentum =I1W1 + I2W2 = 0.171475*39    +0= 6.6875 kgm2/s

2)initial rotational energy =0.5*I1W1^2 + 0.5*I2W2 = 0.5*0.171475*39*39 +0 = 130.4 J

3) anular momentum is conserved,

so (I1+I2)W = 6.6875

   W =6.6875 /(I1+I2) = 6.6875 /( 0.171475 +0.03971)

    = 31.67 rad/s

4) final angular momentum of the rod and disk system = 6.6875 kgm2/s

5)final rotational energy of the rod and disk system = 0.5*( 0.171475 +0.03971) *31.67*31.67

                 = 105.9 J

6) Average torque = i2 * w/t = 0.03971*31.67 /6.5 = 0.1935 Nm

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