Changes in Mechanical Energy for Nonconservative Forces

  • Thread starter Bri
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  • #1
Bri
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I have a problem for the subject in the title. I'm not really sure where to start, and I can't figure it out.

In her hand a softball pitcher swings a ball of mass .250 kg around a vertical circular path of radius 60.0 cm before releasing it from her hand. The pitcher maintains a component of force on the ball of constant magnitude 30.0 N in the direction of motion around the complete path. The speed of the ball at the top of the circle is 15.0 m/s. If she releases the ball at the bottom of the circle, what is its speed upon release?

If someone could point me in the right direction here...
Thanks.
 

Answers and Replies

  • #2
Doc Al
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How much work does she do on the ball? (So what happens to the total mechanical energy?)
 
  • #3
Bri
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Wouldn't the work she does be 0? Since it goes in a circle and ends up back at the starting position?
 
  • #4
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It's a rotational kinematics question. First, with the info given, you can work out angular acceleration, initial angular speed and the angular displacement. After that, just plug them into the most familiar equation in kinematics question to find the final angular speed, which then allow you to find its tangential speed.
 
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  • #5
Doc Al
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Bri said:
Wouldn't the work she does be 0? Since it goes in a circle and ends up back at the starting position?
No. That reasoning would apply to a conservative force, but the force of her hand on the ball is not a conservative force.

In any case, she applies a force over a distance, so work is done.
 
  • #6
Bri
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I don't think I know any of that angular stuff you posted about...
We're supposed to be using Kinetic Energy, Potential Energy, Work, etc. to find the answer.
 
  • #7
Doc Al
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Bri said:
I don't think I know any of that angular stuff you posted about...
We're supposed to be using Kinetic Energy, Potential Energy, Work, etc. to find the answer.
Using energy methods is the easy way to do this problem.
 
  • #8
Bri
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Ok, so work would be 30 * 2pi*.6 = 113.1
I'm still not sure what to do...
I tried
KE(initial) + U(initial) + W(external) = KE(final) + U(final)
.5(.25)(15)^2 + 9.8(.25)(1.2) + 113.1 = .5(.25)v^2 + 9.8(.25)(0)
I solved for v and got 34 m/s, which according to my book is wrong (should be 26.5 m/s)
 
  • #9
Doc Al
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Bri said:
Ok, so work would be 30 * 2pi*.6 = 113.1
I'm still not sure what to do...
I tried
KE(initial) + U(initial) + W(external) = KE(final) + U(final)
.5(.25)(15)^2 + 9.8(.25)(1.2) + 113.1 = .5(.25)v^2 + 9.8(.25)(0)
I solved for v and got 34 m/s, which according to my book is wrong (should be 26.5 m/s)
Your method is fine, but you made a mistake in calculating the work. The work is done over half the circle, not the whole. (Since you are given the speed at the top of the circle.)
 
  • #10
Bri
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Oh, ok!
Thanks so much for the help!
 
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