A person of mass 75 kg stands at the center of a rotating merry-go-round platfor

Question

A person of mass 75 kg stands at the center of a rotating merry-go-round platform of radius 3.5m and moment of inertia 950kg*m^2 . The platform rotates without friction with angular velocity 0.80rad/s . The person walks radially to the edge of the platform.

a)Calculate the angular velocity when the person reaches the edge

b)Calculate the rotational kinetic energy of the system of the platform plus person before the person's walk c)Calculate the rotational kinetic energy of the system of the platform plus person after the person's walk

Explanation / Answer

I final = 950 + mr^2 = 950 + 73*3.5^2 = 1844.25 kg m^2 (a) I1 w1 = I2 w2 950*0.9 = 1844.25*w2 w2 = 0.46 rad/s (b) KE initial = 1/2 I w^2 = 1/2 * 950 * 0.9^2 = 384.75 J (c) KE final = 1/2 I w^2 = 1/2 * 1844.25 * 0.46^2 = 195.12 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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