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Motion of a disc with a particle

  • #1
rock.freak667
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Homework Statement


http://img372.imageshack.us/img372/4379/diagyx7.jpg [Broken]


A uniform disc has centre O,radius a and mass 2m. It is free to rotate in a vertical plane about a fixed horizontal axis through O. A particle of mass m is placed on the highest point of the rough edge of the disc and the system is slightly disturbed so that OP begins to rotate with the particle in conact with the edge. In the subsequent motion, OP makes an angle [itex]\theta[/itex] with the upward vertical (see diagram), For the motion while the partcile remains in contact with the disc without slipping,find [itex]a \ddot{\theta}[/itex] and [itex]a \dot{\theta}^2[/itex] in terms of g and [itex]\theta[/itex]

Show that if the particle remains in contact with the disc,then it begins to slip when

[itex]4\mu cos\theta -sin\theta =2\mu[/itex]
where [itex]\mu[/itex] is the coefficient of friction between the particle and the edge of the disc,

Show that,however large the value of [itex]\mu[/itex],the particle cannot lose contact with the disc before it starts to slip

Homework Equations



Not sure which eq'ns are relevant here

The Attempt at a Solution



Need help on starting. Since the system can rotate,I think I need to use moment of inertia in it somewhere, but the only way to get [itex]a \ddot{\theta}[/itex] from a moment of inertia is to use [itex]\tau =I \alpha[/itex]

But it also talks of the mass slipping,so should I resolve the weight at P into components and find the centripetal acceleration?
 
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Answers and Replies

  • #2
Doc Al
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Need help on starting. Since the system can rotate,I think I need to use moment of inertia in it somewhere, but the only way to get [itex]a \ddot{\theta}[/itex] from a moment of inertia is to use [itex]\tau =I \alpha[/itex]
Yep. So get busy!
But it also talks of the mass slipping,so should I resolve the weight at P into components and find the centripetal acceleration?
Sure. Compare the components of the weight to the friction' force. As far as centripetal acceleration, that's one of the first things they ask you to find. (You'll need it, of course.)
 
  • #3
rock.freak667
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the total moment of inertia is 2ma^2

the torque would be Fasin(theta), where F is the centripetal force? (which is mgsin from the component of the weight)

tangetially, the component is mgcos=mu*R since there is not resultant tangential acceleration.
 
  • #4
Doc Al
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the total moment of inertia is 2ma^2
Good!
the torque would be Fasin(theta), where F is the centripetal force? (which is mgsin from the component of the weight)

tangetially, the component is mgcos=mu*R since there is not resultant tangential acceleration.
Let's start over. If you were to draw a free body diagram of this system, what forces would be on it that produce a torque? (Hint: "Centripetal force" should never appear on a free body diagram.)
 
  • #5
rock.freak667
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Let's start over. If you were to draw a free body diagram of this system, what forces would be on it that produce a torque? (Hint: "Centripetal force" should never appear on a free body diagram.)

well the only forces acting on the particle are the component of weight (mgcos) along the tangent, mgsin acting towards the centre O and the friction uR acting along the tangent but in the opposite direction.
 
  • #6
Doc Al
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well the only forces acting on the particle are the component of weight (mgcos) along the tangent, mgsin acting towards the centre O and the friction uR acting along the tangent but in the opposite direction.
Even simpler. Consider forces acting on the entire system, not just the particle. Only external forces count--so you can forget about friction. (For now.)
 
  • #7
rock.freak667
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Even simpler. Consider forces acting on the entire system, not just the particle. Only external forces count--so you can forget about friction. (For now.)
Well the only other force,I didn't consider, was the weight of the disc,which can also be put into components with 2mgcos in the direction of the tangent and 2mgsin through O
 
  • #8
Doc Al
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Well the only other force,I didn't consider, was the weight of the disc,which can also be put into components with 2mgcos in the direction of the tangent and 2mgsin through O
Careful: Where does the weight of the uniform disk effectively act? Does that weight produce any torque about the axis?
 
  • #9
rock.freak667
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It acts through O,so it produces no torque about O
 
  • #10
Doc Al
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It acts through O,so it produces no torque about O
Right. The weight of the disk acts through its center of mass, which means it produces no torque. The only force producing a torque on this system is the weight of the particle. Use that to find the angular acceleration of the system.
 
  • #11
rock.freak667
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Then resultant torque is [itex]mgasin \theta[/itex]

and so [itex] I \alpha = mgasin\theta[/itex]

[tex]\ddot{\theta}=\frac{mgasin\theta}{2ma^2}[/tex]

[tex]a \ddot{\theta}=\frac{gasin\theta}{2}[/tex]

and for [itex]\dot{\theta}^2[/itex], use the fact that the centripetal force is equal to the components of the weight of particle+ comp. of weight of disc which point toward the centre?
 
  • #12
Doc Al
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Then resultant torque is [itex]mgasin \theta[/itex]

and so [itex] I \alpha = mgasin\theta[/itex]

[tex]\ddot{\theta}=\frac{mgasin\theta}{2ma^2}[/tex]
Good.
[tex]a \ddot{\theta}=\frac{gasin\theta}{2}[/tex]
Correct the typo in that last step.

and for [itex]\dot{\theta}^2[/itex], use the fact that the centripetal force is equal to the components of the weight of particle+ comp. of weight of disc which point toward the centre?
Careful. The centripetal force is the net force pulling the particle towards the center. Weight is only one force with a radial component--there's also a normal force to consider. And the weight of the disk is irrelevant.

In any case, you first need to find the angular speed as a function of theta. Hint: Consider energy conservation.
 
  • #13
rock.freak667
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Well doesn't the normal reaction of the particle, point in the direction towards O ([itex]mgsin\theta[/itex] and the component of the weight of the disc point in the same direction ([itex]2mgsin\theta[/itex]), so the sum of those two provide the centripetal force

making

[tex]m \omega^2 a =mgsin\theta + 2mgsin\theta[/tex]
 
  • #14
Doc Al
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Well doesn't the normal reaction of the particle, point in the direction towards O ([itex]mgsin\theta[/itex] and the component of the weight of the disc point in the same direction ([itex]2mgsin\theta[/itex]), so the sum of those two provide the centripetal force

making

[tex]m \omega^2 a =mgsin\theta + 2mgsin\theta[/tex]
No. Several points:

(1) The centripetal force acting on the particle is the net radial component of the forces acting on the particle. The only forces acting on the particle (that have a radial component) are its weight and the normal force of the disk on the particle.

(2) The radial component of the particle's weight is not [itex]mgsin\theta[/itex]. (Check your angles.)

(3) The weight of the disk is irrelevant. (And, in any case, it acts on the disk, not on the particle.)

In any case, you still need to figure out [itex]\omega[/itex] as a function of [itex]\theta[/itex]. (Use the hint I gave earlier.)
 
  • #15
rock.freak667
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Well using your hint.I took the horizontal line through the base of the disc as having 0 P.E.

and the change in PE from the max height,to the level where P is [itex]mg(a-acos\theta)[/itex]

and so


[tex]\frac{1}{2}I \omega^2=mg(a-acos\theta)[/tex]



I don't think I was supposed to use kinetic energy as well in the conservation, as that wouldn't really help with [itex]\omega[/itex]
 

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