A child is standing on the edge of a merry-go-around that has the shape of a solid disk, as shown in the figure. The mass of the child is 40 kilograms. The merry-go-around has a mass 200 kilograms and a radius of 2.5 meters and it is rotating with an angular velocity of 2.0 radians per second. The child then walk slowly toward the center of the merry-go-around. What will the final angular velocity of the merry-go-around when the child reaches the center? (The size of the child can be neglected).
A. 2.0 rad/s
B. 2.2 rad/s
C. 2.4 rad/s
D. 2.6 rad/s
E. 2.8 rad/s
(GR0177 #89)
Solution:
Angular Momentum: L = Iω
Conservation of Angular Momentum:
∑ Li = ∑ Lf
(Im + Ic)ω = (Im' + Ic')ω'
ω' = ω(Im + Ic)/(Im' + Ic')
Moment inertia of the uniform disk is the same before and after the child moves to the center:
Im = Im' = Im = ½ mmr2 = ½ (200)(2.5)2
ω = 2 rad/s
Initally, moment inertia of the child:
Ic = mcr2 = (40)(2.5)2
When the child reaches the center, r = 0:
Ic' = 0
Thus,
ω' = ω(Im + Ic)/(Im' + Ic')
= (2) [½(200)(2.5)2 + (40)(2.5)2] / [½ (200)(2.5)2]
= (2) [½(200) + (40)] / [½ (200)]
= 280/100
= 2.8 rad/s
Answer: E
ω' = ω(Im + Ic)/(Im' + Ic')
Moment inertia of the uniform disk is the same before and after the child moves to the center:
Im = Im' = Im = ½ mmr2 = ½ (200)(2.5)2
ω = 2 rad/s
Initally, moment inertia of the child:
Ic = mcr2 = (40)(2.5)2
When the child reaches the center, r = 0:
Ic' = 0
Thus,
ω' = ω(Im + Ic)/(Im' + Ic')
= (2) [½(200)(2.5)2 + (40)(2.5)2] / [½ (200)(2.5)2]
= (2) [½(200) + (40)] / [½ (200)]
= 280/100
= 2.8 rad/s
Answer: E
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