Friction and Rotation: Understanding the Ratio of Forces on a Rotating Disk

In summary, the balls are stuck on the disk as it rotates and the centripetal acceleration is 2 due to the friction between the balls.
  • #1
lc99
161
3

Homework Statement


A phonograph record is whiring around at 103 rpm. Two balls of mass 1 kg are sitting on the disk and are at rest with respect to disk. The first ball (1) sits at a radius 5 away from center. The second ball (2) sits at a radius of 10 away from center. What is the ratio of friction forces acting on them? Ratio = (Friction of 1/Friction of 2)

A)0.5
B) 1
C) 2
D) .25
E) not enough info

Homework Equations


T = Ialpha

The Attempt at a Solution


Okay, first,

I know that the disk is rotating at a constant velocity. This means that angular acceleration is 0, thus, the Torque is 0. If the torque is 0 for this disk that is rotating constantly, then that means that net torque is 0 because friction is acting on the balls. Therefore, some force is oppose this friction for net torque to be 0.

However, this doesn't really help me because Torque is just 0? I'm thinking the answer is either E) or could be B) or C) if I am thinking wrong
 
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  • #2
The balls are stuck on the disk as it rotates, so the friction forces on each are enough to keep them from sliding. What is the equation for the centripetal acceleration on an object as a function of the radius?
 
  • #3
berkeman said:
The balls are stuck on the disk as it rotates, so the friction forces on each are enough to keep them from sliding. What is the equation for the centripetal acceleration on an object as a function of the radius?

well,

Ac = v^2/r --> w^2*r

Is Ac the torque?

If it is, then the friction force is twice the ball 2 for ball one because of Radius doubled for ball 1
 
Last edited:
  • #4
lc99 said:
well,

Ac = v^2/r --> w^2*r
Yep. So what's the correct answer to the question then... :smile:
 
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  • #5
berkeman said:
Yep. So what's the correct answer to the question then... :smile:
C) 2! Thanks :D

Unfortunately, i guessed ratio of 1 on my exam out of confusion and panic :(. Atleast I get it now!
 
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  • #6
berkeman said:
The balls are stuck on the disk as it rotates
That's unclear. If glued to the disk, why make them balls, and where does friction come into it?
A feasible interpretation is that the balls are only instantaneously at rest relative to the disc. But if so, we should have been told that there is sufficient friction that the balls do not, at first at least, slide.
Pretty sure the answer would still be 2, though.
 

Related to Friction and Rotation: Understanding the Ratio of Forces on a Rotating Disk

1. What is friction?

Friction is a force that resists the relative motion of two surfaces that are in contact with each other.

2. How does friction affect rotational motion?

Friction can cause a rotational object to slow down or stop due to the resistance it creates against the rotation.

3. What factors affect the amount of friction in a rotating system?

The amount of friction in a rotating system can be affected by the type of surfaces in contact, the force pressing the surfaces together, and the speed of rotation.

4. How can friction be reduced in a rotating system?

Friction can be reduced by using lubricants, such as oil or grease, between the surfaces in contact. Additionally, using smoother surfaces and reducing the force pressing the surfaces together can also decrease friction.

5. Can friction be beneficial in rotational motion?

Yes, friction can be beneficial in certain situations, such as in vehicle brakes or in the grip of tires on the road. It can also be used to control the speed of rotation in machines and equipment.

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