A solid disk of mass m 1 = 9.7 kg and radius R = 0.24 m is rotating with a const

Question

A solid disk of mass m1 = 9.7 kg and radius R = 0.24 m is rotating with a constant angular velocity of = 38 rad/s. A thin rectangular rod with mass m2 = 3.8 kg and length L = 2R = 0.48 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 = 6.8 s to accelerate to its final angular speed with the disk. What average torque was exerted on the rod by the disk?

PLEASE SHOW STEPS

Explanation / Answer

Given that

The mass of the disk (m1) =9.7kg

The radius of the disk (r) =0.24m

The moment of inertia of the disk is ID =(9.7)(0.24)2/2 =0.279kg.m2

The mass of the rod (m2) =3.8kg

Length of the rod (L) =0.48m

The moment of inertia of the rod (IR) =m2L2/12 =3.8kg*(0.48)2/12 =0.0729kg.m2

1)

The initial angualr speed of the rod is wR =0

The inital angualr speed of the disk is wD =38rad/s

The initial angualr momentum is

Linitial =IDwD +IRwR =(0.279)(38)+0=10.602kg.m2rad/s

2

The initial rotational energy is

Kinitial =(1/2)IDwD2 +(1/2)IRwR2

=0.5*(0.279)(38)2 + 0 =201.438J

3.

From the conservation of angualr momentum the final angular speed is

w =Linitial/ID+IR =10.602kg.m2rad/s/(0.279 +0.0729)=30.127rad/s

4.

The final angualr speed is given by

L final = (ID+IR)w =(0.279 +0.0729)*30.127 =10.601kgm2rad/s

5.

The final rotational energy is

KEfinal =(1/2)(ID+IR)w2 =0.5*(0.279+0.0729)(30.127)2 =159.698J

6.

The torque on the rod due to the disk is given by

t =IR(w-wR/t) =(0.0729)(30.127-0/6.8) =0.323N.m

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