A solid disk of mass m 1 = 9.8 kg and radius R = 0.19 m is rotating with a const

Question

A solid disk of mass m1 = 9.8 kg and radius R = 0.19 m is rotating with a constant angular velocity of ? = 32 rad/s. A thin rectangular rod with mass m2 = 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 = 5.8 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)

initial angular momentum of the rod and disk system = 1/2MR^2*w = 0.5*9.8*0.19*0.19*32 =5.66048 kg-m2/s

2)

initial rotational energy of the rod and disk system = 1/2*1/2MR^2*w^2 = 0.5*0.5*9.8*0.19*0.19*32*32 =90.56768 J

3)

we know that Id = (1/2) M R^2 = 0.5*9.8*0.19*0.19 = 0.17689

and we can look up Ir = (1/12) m L^2 = 3 * 0.38 *0.38 /12 =0.0361

so add Id + Ir = 0.17689+0.0361 =0.21299

we know that

I1*w1 = I2*w2

w2 = I1*w1 / I2 = 0.17689*32 / 0.21299 = 26.576

4) the final momentum will be 5.66048 kg-m2/s

5)KE2 = 0.5*I2*w2^2 =0.5*0.21299*26.576*26.576 =75.2156

6)

Irod = (1/12)m*L^2 = 0.0361 kg?m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.