Static Friction in Circular Motion

Click For Summary

Homework Help Overview

The discussion revolves around a problem involving static friction in circular motion, specifically analyzing forces acting on an object moving in a circular path with given radius and coefficient of static friction.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the acceleration and period of revolution using the forces acting on the object, while questioning the role of mass and the direction of the frictional force in a free-body diagram.

Discussion Status

Some participants affirm the calculations presented by the original poster, while also prompting further exploration of the direction of the net force and frictional force in circular motion. There is an ongoing clarification regarding the implications of mass on the results.

Contextual Notes

The original poster expresses uncertainty about the frictional force's direction and its impact on the calculations, indicating a need for deeper understanding of the concepts involved.

Softwarm
Messages
2
Reaction score
1
Homework Statement
You are sitting on the edge of a horizontal disk (for example, a playground merry-go-round) that has radius 3.00 m and is rotating at a constant rate about a vertical axis. The coefficient of static friction between you and the surface of the disk is 0.280

A) What is the minimum time for one revolution of the disk if you are not to slide off? Express your answer with the appropriate units.

B) Your friend's weight is half yours. If the coefficient of static friction for him is the same as for you, what is the minimum time for one revolution if he is not to slide off? Express your answer with the appropriate units.
Relevant Equations
a = 4*π^2*r / T^2
A) So we are given the radius and the coefficient of static friction as 3.0 m and 0.28 respectively. I know that in the vertical direction the only forces acting are the normal force and the gravitational force. Therefore, the normal force is equal to mg because net force is equal to 0, due to no vertical acceleration.

I can find the force of static friction, FS = FN * μS
Net force in the horizontal direction, ma = FS, the only force is the force of friction
ma = mg * μS
a = g * 0.28
a = 2.744 m/s2

I'm not entirely sure where the frictional force would point on a free-body diagram but, I can use the acceleration to solve for the time for one revolution (T) with the equation mentioned above.

2.744 = 4 * π2 * (3) / T2
T = √(4 * π2 * (3) / 2.744)
T = 6.57 s

B) For this part, I don't think the differences in mass will have an affect on the time for one revolution.
When solving for the acceleration,
0.5m * a = 0.5m * g * μS
The 0.5m will just cancel out and we'll be left with the same acceleration and the same time for one revolution.
T = 6.57 s

Is my answer correct? I'm not sure if there is anything I missed or if I used my equations incorrectly.
 
Physics news on Phys.org
Welcome to PF!

Your work looks good. I think your answers for (a) and (b) are correct.

Softwarm said:
I'm not entirely sure where the frictional force would point on a free-body diagram
Whenever an object moves in a circle at constant speed, what is the direction of the net force acting on the object? Therefore, what must be the direction of the friction force in this problem?
 
  • Like
Likes   Reactions: Softwarm
Softwarm said:
FS = FN * μS
Only in the limiting case, where it is on the verge of slipping (as here).
 
  • Like
Likes   Reactions: Softwarm
TSny said:
Welcome to PF!

Your work looks good. I think your answers for (a) and (b) are correct.Whenever an object moves in a circle at constant speed, what is the direction of the net force acting on the object? Therefore, what must be the direction of the friction force in this problem?

Ahh I see, so it would point towards the center. Thanks.
 
  • Like
Likes   Reactions: TSny

Similar threads

Replies
61
Views
4K
Replies
9
Views
2K
Replies
43
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
11
Views
1K
Replies
4
Views
1K
Replies
42
Views
3K
Replies
24
Views
3K
Replies
6
Views
3K