Friction as centripetal force....problem with the mass

In summary: Are they both resting on a flat rotating surface, perhaps? What are the shapes and widths of the objects, and how far are their mass centres from the axis?
  • #1
Geo louv
10
1

Homework Statement



The problem is that if I have a 5kg mass and 1 kg mass at the same radius with the same μ...and I increase slowly the angular speed ω...then the first mass that will stop the circular motion is the one of the 5 kg mass because needs more centripetal force.Can someone say that this is not true and the two objects will stop at the same time the circular motion because ofthe equations T=m(u^2)/r μΝ=m(u^2)/r μ(mg)=m(u^2)/r. μg=(u^2)/r. ?
2. Homework Equations

The Attempt at a Solution

 
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  • #2
I have big problem...because if you see the equation μg=u^2/r you can assume that the two objects will leave at the same time...
 
  • #3
Geo louv said:
.Can someone say that this is not true and the two objects will stop at the same time
This is not true.
 
  • #4
Yes if you see the equation F=mω^2R...then you will come to the conclusion that the one with the bigger mass will stop first to moving circular because need more centripetal force.....Yes but also at the bigger mass I have bigger friction...T=μΝ. Τ=μ(mg)...so why two masses are not stopping the circular motion at the same time??
 
  • #5
They should stop at the same time.
 
  • #6
No...if you do the experiment the bigger mass stop first the circular motion
 
  • #7
Geo louv said:

Homework Statement



The problem is that if I have a 5kg mass and 1 kg mass at the same radius with the same μ...and I increase slowly the angular speed ω...then the first mass that will stop the circular motion is the one of the 5 kg mass because needs more centripetal force.Can someone say that this is not true and the two objects will stop at the same time the circular motion because ofthe equations T=m(u^2)/r μΝ=m(u^2)/r μ(mg)=m(u^2)/r. μg=(u^2)/r. ?
2. Homework Equations

The Attempt at a Solution

This problem is not well defined. You haven't given enough details.

From the sparse description, I don't see how centripetal force is involved at all.
 
  • #8
image.jpg
 
  • #9
But m1 has also bigger friction...because of the bigger mass...so they must leave at the same time
 
  • #10
Geo louv said:
No...if you do the experiment the bigger mass stop first the circular motion
Please describe the experiment in detail. Are they both resting on a flat rotating surface, perhaps? What are the shapes and widths of the objects, and how far are their mass centres from the axis?
 
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Likes Geo louv
  • #11
Haruspex you are GENIUS...OF course if it is a float rotating surface...they will stop rotation at the same time...,but if it's not...and the maximum Ts is the same then...the bigger mass will stop rotation first
 
  • #12
Geo louv said:
Haruspex you are GENIUS...OF course if it is a float rotating surface...they will stop rotation at the same time...,but if it's not...and the maximum Ts is the same then...the bigger mass will stop rotation first
Please describe the experiment in detail.
 

Related to Friction as centripetal force....problem with the mass

1. How does friction act as the centripetal force in a circular motion?

Friction acts as the centripetal force in a circular motion by providing a force that is perpendicular to the velocity of the object and towards the center of the circle. This force is necessary to keep the object moving in a circular path.

2. What causes the frictional force to act as the centripetal force?

The frictional force acts as the centripetal force due to the interaction between the surface of the object and the surface it is moving on. This interaction creates a force that opposes the motion of the object and acts towards the center of the circle.

3. Does the mass of the object affect the amount of frictional force acting as the centripetal force?

Yes, the mass of the object does affect the amount of frictional force acting as the centripetal force. The greater the mass of the object, the greater the frictional force needed to keep it moving in a circular path. This is because a larger mass requires a greater force to change its direction of motion.

4. How can the problem of friction as the centripetal force be reduced?

The problem of friction as the centripetal force can be reduced by using smoother surfaces, such as lubricated materials, to reduce the frictional force. Additionally, reducing the speed at which the object is moving can also decrease the amount of frictional force needed to maintain circular motion.

5. Can friction ever be completely eliminated as the centripetal force?

No, friction cannot be completely eliminated as the centripetal force. Even with smooth surfaces and low speeds, there will always be some amount of friction present. However, it can be minimized to a point where it has minimal impact on the motion of the object.

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