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An object on an inclined plane |
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| Aug10-11, 07:04 AM | #1 |
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An object on an inclined plane
1. The problem statement, all variables and given/known data
A homogenous thin cylinder is put on an inclined plane (angle of elevation is α). The coefficient of friction is [tex] \mu [/tex] . What is the speed of the centre of the mass of the cylinder and it's angular velocity at distance l from the start of motion. 2. Relevant equations [tex] F=ma [/tex] [tex] I=mR^2/2 [/tex] [tex] v=at [/tex] 3. The attempt at a solution I drew a free body diagram and found out that the acceleration is [tex] a=g(sinα-\mu\cosα) [/tex] and found the velocity [tex] v=\sqrt{2lg(sinα-\mu\cosα)} [/tex] The problem is, i can't find the angular velocity in any way. |
| Aug10-11, 07:12 AM | #2 |
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I presume this is pure "rolling", no sliding, motion. In one complete rotation, then, the cylinder will have moved a distance equal to the circumference of the cylinder. Since you know the linear speed of the center axis, you know the time it will take the cylinder to move that far. The angular speed is [itex]2\pi[/itex] radians divide by that time.
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| Aug10-11, 07:12 AM | #3 |
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| Aug10-11, 08:22 AM | #4 |
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An object on an inclined plane
[tex]N=mgcos\alpha[/tex]
[tex]ma=mgsin\alpha-F[/tex] [tex]F=\mu N[/tex] and from here [tex]a=g(sinα−μcosα)[/tex] and the distance l is [tex]l=at^2/2[/tex] from here [tex]t=\sqrt{2l/a}[/tex] and velocity is [tex]v=at=\sqrt{2lg(sinα−μcosα)}[/tex] But that would mean the cylinder is sliding i guess. I'm completely lost now. |
| Aug10-11, 08:41 AM | #5 |
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(Don't forget Newton's 2nd law for rotation.) |
| Aug10-11, 08:56 AM | #6 |
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So
[tex] mg sin \alpha - F = ma [/tex] [tex] FR=mR^2\beta/2 [/tex] [tex]a=\beta R[/tex] and from here [tex]a=2gsin\alpha /3[/tex] and i can find the translational velocity, but what about the angular velocity? it seems i will need the radius, which is not given |
| Aug10-11, 09:07 AM | #7 |
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(If this is a textbook problem, what book is it from?) |
| Aug10-11, 09:31 AM | #8 |
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It's not a textbook problem. However, thanks a lot for your help ! :)
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