A simple conservation of energy problem, with rotation (bowling ball).
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A simple conservation of energy problem, with rotation (bowling ball).

[From: ] [author: ] [Date: 11-04-28] [Hit: ]
And for part e, you just apply conservation of energy and solve for d ?Theres no height right? Because the bowling ball is thrown and lands on the ground. Can I ignore the potential energy.-The initial total energy TE = KE = 1/2 mv^2 before striking the alley.......
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I already did parts a and b.
I have questions about parts c -e. I need to make sure I'm interpreting these parts correctly.

For part c, you just add the initial KE and the final KE right.

For part d, do you apply the work energy theorem?

And for part e, you just apply conservation of energy and solve for d ?

There's no height right? Because the bowling ball is thrown and lands on the ground. Can I ignore the potential energy.

-
The initial total energy TE = KE = 1/2 mv^2 before striking the alley. m is the ball mass and v is its impact speed.

This is degraded by the work done WE = kmgS by the friction as the ball slips along the alley a distance S. k is the friction coefficient (I don't do mus).

So the ke when the ball starts to spin is ke = KE - WE = 1/2 mu^2 (1 + 2/5); where u is both the tangential speed and the linear (CG) speed of the ball with no slippage. The 2/5 factor comes from the angular kinetic energy and the moment of inertia.

As you started the energy audit just prior to impact at height h = 0 on the alley, you need not factor in GPE as there is no difference in heights along the alley (we hope although I've rolled rocks on some very uneven alleys in my day).

NOTE: As you frequent Answers a lot, it would be helpful if you copy and pasted the questions in your question. I was unable to do that at your cited source. So that's why I did not answer each question specifically. Thanks.
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