Rotational Dynamics of a solid uniform disk

  • Thread starter rwx1606
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  • #1
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Homework Statement


A solid uniform disk of mass 21.0 kg and radius 85.0 cm is at rest flat on a frictionless surface. A string is wrapped around the rim of the disk and a constant force of 35.0 N is applied to the string. The string does not slip on the rim. How much string has unwrapped from around the rim if the disk travels 7.3m?

The Attempt at a Solution


The amount of string unwrapped turns out to be 2x the distance traveled. I can't understand why this is so. (Conceptually) Does the disk not translate in the beginning while some amount of string is unwrapping? My gut intuition is the amount of string unwrapped equals the distance traveled by the disk. Can someone explain why this isn't so?
 

Answers and Replies

  • #2
alphysicist
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Hi rwx1606,

Homework Statement


A solid uniform disk of mass 21.0 kg and radius 85.0 cm is at rest flat on a frictionless surface. A string is wrapped around the rim of the disk and a constant force of 35.0 N is applied to the string. The string does not slip on the rim. How much string has unwrapped from around the rim if the disk travels 7.3m?

The Attempt at a Solution


The amount of string unwrapped turns out to be 2x the distance traveled. I can't understand why this is so. (Conceptually) Does the disk not translate in the beginning while some amount of string is unwrapping? My gut intuition is the amount of string unwrapped equals the distance traveled by the disk. Can someone explain why this isn't so?


The amount of string unwrapped would be equal to the distance travelled if the disk were rolling.

However, this disk is sliding, and so that will not be true. In particular, the linear and angular accelerations will not be related by [itex]a = r \alpha[/itex]; you need to find them indepedently. Does this help?
 
  • #3
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Yes, that does help. However, what does it mean to for the disk to slide? Is the angular velocity less than the velocity of the center of mass? or is it greater than the center of mass? I know that if it's slipping then the relationship v=rw does not hold. Another related question is, does a rolling object only roll without slipping if ONLY static friction is providing the torque?
 
  • #4
alphysicist
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Yes, that does help. However, what does it mean to for the disk to slide? Is the angular velocity less than the velocity of the center of mass? or is it greater than the center of mass? I know that if it's slipping then the relationship v=rw does not hold. Another related question is, does a rolling object only roll without slipping if ONLY static friction is providing the torque?

From your questions, I think there's a chance you might not understood my last post. Just to be clear, this disk is not rolling or sliding or skidding like a wheel might do; it's is sliding on its face like an air hockey puck.

So you don't need to relate the linear and angular acceleration; just draw a force diagram. Then apply Newton's law for forces, and Newton's law for torques. What do you get?
 
  • #5
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Sorry. I meant in general for a rolling object and not this particular problem with the disk.
 
  • #6
alphysicist
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Sorry. I meant in general for a rolling object and not this particular problem with the disk.

In general the object can either rotate faster or slower than you would expect from [itex]v=r\omega[/itex]. An example of a faster rotation would be a race car taking off too fast. The tires can spin very fast, but the car is not moving forwards as much as you would expect from the equation.

The opposite example could be a bowling ball. At the beginning of the lane it would (normally) be rotating slower than you might expect from the equation.
 

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