WEBVTT
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This is Giancoli Answers
with Mr. Dychko.
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As this person walks from the middle
of the merry-go-round to the edge,
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there is no external net torque
acting on the system
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so the angular momentum is staying constant.
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Any forces involved in moving this person
from the middle to the edge
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are internal to the system.
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So that means moment of inertia final times
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final angular velocity equals
moment of inertia initial
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times angular velocity initial.
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So the moment of inertia final after
the person moves to the edge
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is gonna be the moment of inertia
of the merry-go-round
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plus the moment of inertia of the person
which we'll treat as a single point mass
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going around in a circle of radius r—
radius of the merry-go-round—
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so that's mass person times radius
merry-go-round squared.
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And the initial angular momentum is just
that of the merry-go-round
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because since the person is in the middle,
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their radius or the distance from
the axis of rotation is zero
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so m pr squared will be zero
in this case initially.
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So we can solve for ω f by dividing
both sides by
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I M plus m pr squared
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and then we substitute in some numbers
and we get
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that the final angular velocity is
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820 kilogram meter squared per second—
moment of inertia of the merry-go-round—
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times—its initial angular velocity of—
0.95 radians per second
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and divide that by 820 plus
75 times 3 squared
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and that gives 0.52 radians per second
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will be the angular velocity after
the person moves to the edge.
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So because the moment of inertia
has increased
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when they move to the edge,
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their angular velocity has to compensatingly
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decrease by the same factor by which
the moment of inertia increased.
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in order for this product of
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moment of inertia and angular velocity
to remain the same, before and after.
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Okay!
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So kinetic energy—before and after;
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initially, the person's at the middle so
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the moment of inertia just that
of the merry-go-round
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so we have kinetic energy initial is
one-half times
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820 kilogram meter squared per second
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times 0.95 radians per second squared
which gives 370 joules.
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And then after the person moves to the edge,
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the thing that changes is the formula
for moment of inertia now is
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that of the merry-go-round plus
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the contribution of the person on the edge
and then times their
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angular velocity squared which is now
the different one
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our answer to part (a)
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the slower angular velocity when
the person reaches the edge.
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So one-half times 820 plus 75 times 3 meters
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times 0.52107 radians per second squared
gives us 202.29566
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and we'll need two significant
figures in that.
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So we could write 200 but that
would be ambiguous
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as to how many significant figures that had
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so I have to write 2.0 times 10 to
the 2 joules, there we go!