A man stands on a platform that is rotating (without friction) with an angular s

Question

A man stands on a platform that is rotating (without friction) with an angular speed of 1.6 rev/s; his arms are outreached and he holds a weight in each hand. The rotational inertia of the system of man, weights, and platform about the central axis is 13.00 kg m2. If by moving the weights the man decreases the rotational inertia of the system to 6.24 kg m2, what is the resulting angular speed of the platform?(rad/s)

What is the ratio of the new kinetic energy of the system to the original kinetic energy?

Explanation / Answer

Key concept : angular momentum co nservation

I*W = i*w

13*1.6 = 6.24*w

w= 13*1.6/6.24 = 3.33 rev/s

w = 3.33 *2pi /1 = 20.92 rad/s ..........answer

2)

Kef /Ke i = [0.5*6.24*20.92^2] / [ 0.5* 13 * ( 1.6*2pi)^2] = 2.07 .......answer

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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