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A 150 N sphere 0.20 m in radius rolls, without slipping 6.0 m down a ramp that is inclined at 34° with the horizontal. What is the angular speed of the sphere at the bottom of the hill if it starts from rest?
(rad/s)

2007-11-11 14:41:25 · 1 answers · asked by Anonymous in Science & Mathematics Physics

1 answers

The moment of inertia of the sphere is
2*m*r^2/5
assuming it is solid

The KE gained by descending the hill is
m*g*h

150*6*sin(34)
503 J

The KE of the sphere has two parts
.5*m*v^2
and
.5*I*w^2
since v=w*r
when there is no slipping

503=.5*m*w^2*r^2*(2/5+1)

w=34 rad/sec

j

2007-11-13 05:44:53 · answer #1 · answered by odu83 7 · 0 0

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