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Rolling friction

  • Thread starter khangaroo
  • Start date
  • #1
2
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Homework Statement



A disk of mass 6 kg and outer radius 60 cm with a radial mass distribution (which may not be uniform) so that its moment of inertia is [tex]\large_{{\frac{2}_{7}}mR^2}[/tex]. The disk is rotating at angular speed 7 rad/s around its axis when it touches the surface, as shown. The disk is carefully lowered onto a horizontal surface and released at time [tex]\large_{t_{0}}[/tex] with zero initial linear velocity along the surface. Assume that when the disk lands on the surface it does not bounce. The coefficient of friction between the disk and the surface is 0.08.

The kinetic friction force between the surface and the disk slows down the rotation of the disk and at the same time gives it a horizontal acceleration. Eventually, the disk's linear motion catches up with its rotation, and the disk begins to roll (at time [tex]\large{t_{rolling}}[/tex]) without slipping on the surface.

Once the disk rolls without slipping, what is its angular speed? The acceleration of gravity is 9.8 m/s2.

Homework Equations



[tex]\omega=\omega_{0}-\alpha{t}[/tex]
[tex]\tau=I\alpha[/tex]

The Attempt at a Solution



torque = inertia * angular acceleration
force of friction * radius of disk = inertia * angular acceleration
[tex]\alpha={\frac{fR}_{I}}={{\frac{\mu{mgR}}_{{\frac{2}_{7}}mR^2}}}={\frac{7}_{2}}{\frac{\mu{g}}_{R}}[/tex]
[tex]\omega=\omega_{0}-{\frac{7}_{2}}{\frac{\mu{g}}_{R}}{t}[/tex]

I'm not sure if this is right so far, and I don't know how to solve for t.

Thanks!
 

Answers and Replies

  • #2
430
2
You have written the equations for rolling but not for translation. Why? That should give the time and final velocity.
Alternatively, you could have conserved angular momentum about a point on the floor.
 

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