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Need help on physics question..keep getting it wrong. LINK INCLUDED

[From: ] [author: ] [Date: 12-06-27] [Hit: ]
33rad/sThe angular momentum of the system is then L=Iw=23.5kgm^2*2.33rad/s=54.76kgm^2/sThe rotational kinetic energy of the system is K = 1/2Iw^2 = 1/2*23.5kgm^2*(2.33rad/s)^2 = 63.......
The rotational inertia of the system is I=40kg*(0.588m)^2 + 57kg*(0.412m)^2=23.5kgm^2
The angular velocity of the both object is w=1 rev/2.7s=(2pi)radians/2.7s=2.33rad/s

The angular momentum of the system is then
L=Iw=23.5kgm^2*2.33rad/s=54.76kgm^2/s
The rotational kinetic energy of the system is
K = 1/2Iw^2 = 1/2*23.5kgm^2*(2.33rad/s)^2 = 63.8J

The bowling ball has both translational and rotational kinetic energy. The mass and speed are given so the translational KE should be easy to find (1/2m*v^2). But to find the rotational KE the rotational inertia (given) and the angular velocity are needed (not given). Angular velocity can be calculated by change in angle/time. Since the ball rolls without slipping we know that it makes a complete rotation (2pi radians) when it has traveled a distance equal to a circumference (2pi*r). Since we know the speed and radius we can find the time it takes to make a complete rotation and put it into angular velocity equation.
Time, t, =d/v=2pi*r/v=2pi*0.45m/4.4m/s=0.643s.
Angular velocity, w, = radians/time= 2pi/0.643s= 9.77 rad/s

So now we can find total kinetic energy
KE=KE(translational)+KE(rotational)
=1/2m*v^2+1/2I*w^2
=1/2 (4.9kg)(4.4m/s)^2 + 1/2 (0.414736kgm^2)*(9.77rad/s)^2
=47.43J+19.79J
=67.22J

Of course this looks like a quest problem which can be sensitive to significant figures.
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