ap physics rolling, angular momentum, torque?

Question

A bowling ball encounters a 0.76 m vertical rise on the way back to the ball rack, as the drawing illustrates. Ignore frictional losses and assume that the mass of the ball is distributed uniformly. If the translational speed of the ball is 4.80 m/s at the bottom of the rise, find the translational speed at the top.

Answer 1

Energy in the ball is stored in two ways:
translational kinetic energy and rotational kinetic energy
translational kinetic is .5*m*v^2
rotational kinetic is
.5*I*w^2

For a solid sphere, I=2/5 m*r^2

the angular velocity is related to the translational velocity as
w=v/r


At the beginning of the ramp the total energy of the system is
.5*m*4.8^2+
.5*2/5*m*r^2*4.8^2/r^2

simplifying
.5*m*4.8^2*(7/5)

the rise causes a transformation of kinetic energy to potential energy as
m*g*.76

so the final energy can be expressed as
.5*m*v^2*(7/5)+m*g*.76=
.5*m*4.8^2*(7/5)

subtract m*g*.76 and multiply by 2 and divide by m on both sides
v^2*(7/5)=4.8^2*(7/5)-2*g*.76

multiply by 5/7
v^2*=4.8^2-2*g*.76*(5/7)

using g=9.81
v^2=12.4
v=3.52 m/s

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