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 1.7 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 11.00 kg m2. If by moving the weights the man decreases the rotational inertia of the system to 4.62 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

initial angular velocity is wi = 1.7 rev/s

final angular velocity is wf = ?

initial moment of inertia is I1 = 11 kg-m^2

final moment of inertia is I2 = 4.62 kg-m^2

then using law of conservation of angular momenmtum

I1*w1 = I2*w2

w2 = wf = (I1*w1)/I2 = (11*1.7)/(4.62) = 4.04 rev/sec = 4.04*2*3.142 = 25.38 rad/s

------------------------------------------------------------------------------

KE_new /KE_original = (4.62*4.04^2)/(11*1.7^2) = 2.37

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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