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After you pick up a spare, your bowling ball rolls without slipping back toward the ball rack with a linear speed of v = 3.22 m/s (Figure 10-24). To reach the rack, the ball rolls up a ramp that gives the ball a h = 0.50 m vertical rise. What is the speed of the ball when it reaches the top of the ramp?
m/s

2007-12-05 12:23:03 · 1 answers · asked by Amber G 2 in Science & Mathematics Physics

1 answers

this will be solved using energy conservation.
The energy of the rolling ball has two components:
translational
.5*m*v^2
and rotational
.5*I*ω^2
where ω is the angular speed and ω=v/r
I is the moment of inertia.
Assuming the bowling ball to be a solid sphere
I=2*m*r^2/5

the gain in PE is m*g*h

setting up the equations, note the m will divide out
3.22^2*(.5+1/5)-9.81*0.5=v^2*(.5+1/5)
solve for v

v=1.83 m/s

j

2007-12-06 04:27:33 · answer #1 · answered by odu83 7 · 0 0

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