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rolling without slipping - too many equations |
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| May29-12, 05:51 PM | #1 |
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rolling without slipping - too many equations
1. The problem statement, all variables and given/known data
I think I have too many equations for unknowns for rolling without slipping. Suppose a wheel is pushed and left rolling along the ground without slipping. 2. Relevant equations T = I*alpha, where alpha is the rotation about the wheel's centroid and I is the moment of inertia. F = ma. omega = r*v, where omega is the rotational speed, r is the wheel radius and v is the translational speed. The rolling without slipping condition. From this, alpha = r*a. T = r*F, where F is friction. 3. The attempt at a solution I*alpha = r*F, so alpha = r*F/I. For rolling without slipping, a = r^2 * F/I. But from Newton's second law, a = F/m. What am I doing wrong. Thanks. |
| May29-12, 07:35 PM | #2 |
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There are 3 'F' here not one.
F force apllied Fnet=ma Ffriction=Iα Fnet=F apllied±Ffriction=ma |
| May30-12, 07:46 AM | #3 |
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| May30-12, 07:56 AM | #4 |
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rolling without slipping - too many equations
So the wheel is just rolling along with a = 0, since there's no longer a force pushing it?
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| May30-12, 08:01 AM | #5 |
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Without external forces, the wheel will keep on rolling forever at initial rotational and translational velocity.
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| May30-12, 08:24 AM | #6 |
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What, according to you, would happen if someone rotates the wheel clockwise and
A) there is no friction B) there is friction. In this case what will be direction of friction? |
| May30-12, 08:33 AM | #7 |
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The constraint is rolling without slipping.
Then friction as the torque for rotation. A. Will rotate forever like the satellite. B. The friction is anticlockwise. |
| May30-12, 10:27 AM | #8 |
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| May30-12, 10:30 AM | #9 |
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| May30-12, 10:42 AM | #10 |
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| May30-12, 11:45 AM | #11 |
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| May30-12, 02:03 PM | #12 |
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You have to be careful and distiguish between 2 types of friction....sliding friction and rolling friction. When something like a snooker ball is first struck the ball slides over the surface and sliding friction causes the balls linear speed to decrease.
The frictional force acting on the ball produces a turning effect which makes the ball rotate. At the point of contact with the table the forward velocity will decrease to a value, v, and the (backwards) rotation velocity will increase to a velocity, v. At this point there is no relative motion between the point of contact on the ball and the point of contact on the table so sliding friction will become zero. The ball will then continue to roll with no sliding (slipping). The link between linear velocity and angular velocity when this happens is v = ωr There is now only rolling friction which is very small compared to sliding friction |
| May30-12, 02:26 PM | #13 |
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| May30-12, 02:33 PM | #14 |
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Your derivation takes "F" as some force acting on the wheel. What's that force? The applied force? But that's zero, since you stopped pushing. Friction? There is none. (I'm talking static friction, not rolling friction.) |
| May30-12, 05:02 PM | #15 |
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| May30-12, 11:10 PM | #16 |
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Rolling friction is static friction.
Just like a block on the a rough table, the static fraction is equal or less to the force applied. When no force applied, static friction is equal to zero. When no force applied on rotating object without slippage then slipping friction is zero. |
| May31-12, 05:12 AM | #17 |
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Actual rolling friction is a bit more complicated and is the force that will eventually stop the wheel as it rolls along the horizontal surface. |
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