Bowling ball rolling question

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Homework Statement


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Homework Equations



Wnon-conservative forces = ΔEnergy

The Attempt at a Solution


I understand that I am to solve for the D in Wnon-conservative forces since I know the friction force. I am just having trouble understanding what the initial/final kinetic rotational/translational energy of the system. It is saying derive an expression of distance BEFORE it begins rolling forward without slipping, I know that if a sphere is rolling forward without slipping, it has both kinetic translational and rotational energy, so if it's saying before this happens, are they saying that the initial conditions only include translational kinetic energy?
 
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Answers and Replies

  • #2
BruceW
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it says it is released with translational velocity 'v' and angular velocity 'w', and they want you to give your answer in terms of these parameters. so you should not assume that there is initially only translation. just assume that the initial rotation is not enough to prevent 'slipping'. And then they want you to work out when the ball stops slipping.
 
  • #3
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Ok, so when the ball is released, it both has kinetic rotational and translational energy and since it is released with a back spin, that means it is rotating opposite the direction of motion so it would be negative, I am thinking that would be my initial conditions. Now for the final conditions, what does it mean for when the ball to stop "slipping"?
 
  • #4
haruspex
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what does it mean for when the ball to stop "slipping"?
If a ball radius r is rolling on a stationary surface, what's the relationship between its angular velocity and its linear velocity?
 
  • #5
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If a ball radius r is rolling on a stationary surface, what's the relationship between its angular velocity and its linear velocity?
it is V/R = w , but I still don't understand whether the final conditions has rotational or translational energy, BruceW suggested to work out the problem when the ball stops slipping,will that mean the final will just include a rotational kinetic energy?
 
  • #6
rcgldr
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when the ball stops slipping,will that mean the final will just include a rotational kinetic energy?
That is a possible outcome, in addition to transitioning into rolling without sliding. Using an energy approach makes this problem more complicated than it needs to be.

What you have is a kinetic (dynamic, sliding) friction force that results in linear deceleration of the bowling ball, and that same kinetic friction force times the radius of the ball results in a torque that in turn produces angular acceleration (decleration from back spin, then acceleration into forwards spin). Given these factors, there's some point in time and distance where the negative of the angular velocity times the radius equals the linear velocity of bowling ball (assuming ball is moving left to right as shown in the diagram and that clockwise rotation is negative angular velocity) , in which case the bowling ball has transitioned into rolling (without slipping) or the bowling ball has completely stopped.
 
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  • #7
ehild
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it is V/R = w , but I still don't understand whether the final conditions has rotational or translational energy, BruceW suggested to work out the problem when the ball stops slipping,will that mean the final will just include a rotational kinetic energy?
If the ball rolls without slipping its KE has both a translational part : the KE of the centre of mass, 1/2 mv2, and a rotational part 1/2 I ω2. KE=KE(translation) + KE(rotation). But rolling means that the speed of the CM is equal to the linear speed of the perimeter : v=rω. You can express the final KE in terms of either the speed of translation or the angular speed of rotation. If the ball stops its rotation also stops. Translation and rotation coexist as soon as the ball starts rolling.

ehild
 
  • #8
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If the ball rolls without slipping its KE has both a translational part : the KE of the centre of mass, 1/2 mv2, and a rotational part 1/2 I ω2. KE=KE(translation) + KE(rotation). But rolling means that the speed of the CM is equal to the linear speed of the perimeter : v=rω. You can express the final KE in terms of either the speed of translation or the angular speed of rotation. If the ball stops its rotation also stops. Translation and rotation coexist as soon as the ball starts rolling.

ehild
I can see what you're saying for the terms for the final KE, you're saying I can express it in either terms of either translation or angular, but I still don't understand why I can express the final terms as just one kinetic energy term? cause from what I am understanding is I can just express the final kinetic energy term as

1/2m(rw)^2 or 1/2I(v/r)^2
 
  • #9
ehild
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I can see what you're saying for the terms for the final KE, you're saying I can express it in either terms of either translation or angular, but I still don't understand why I can express the final terms as just one kinetic energy term? cause from what I am understanding is I can just express the final kinetic energy term as

1/2m(rw)^2 or 1/2I(v/r)^2
No. It is KE = 1/2mv2+1/2I(v/r)2. Plug in the expression for I and simplify.
 
  • #10
BruceW
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yeah. the key is freshcoast's post 5 (in response to post 4): V/R=W for a ball rolling on a stationary surface. This is the condition for "when the ball stops slipping". Also, as ehild just posted, the final KE is a sum of translational and rotational terms. Also, the ball is not necessarily initially rotating backwards. It just needs to be rotating forwards at an angular velocity of less than V/R.
 
  • #11
rcgldr
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Note that KE is not conserved since kinetic friction is converting some of the energy into heat. I think that the problem would be easier to solve if you focused instead on forces and linear accelerations, plus torques and angular accelerations. After you've solved the problem, you can then determine how much energy was lost to kinetic friction (converted to heat).
 

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