# Circular Motion - friction and tension problems

• thebrandon
In summary: Tcos53.13)Substituting in the value for T gives me:(v^2)/.75 = (1.25*9.8cos53.13 + (3.03m / .75)cos53.13)Simplifying..(v^2)/.75 = (15.4125 + 4.04m)cos53.13I'm stuck here... any ideas?The tension in the lower cord is the other force acting on the mass. You should be able to write an equation with that and the
thebrandon

## Homework Statement

A curve with a 125 radius on a level road is banked at the correct angle for a speed of 20. If an automobile rounds this curve at 30 , what is the minimum coefficient of static friction between tires and road needed to prevent skidding?

## Homework Equations

F=ma
centripetal acceleration = (v^2)/r

## The Attempt at a Solution

My teacher likes us to leave the x and y axes in there standard places, and create new axes in the parallel and perpendicular directions. In this case, the parallel axis would be the track, and the perpendicular axis would be perpendicular to the track.

I separated all forces into their parallel and perpendicular components:
Parallel gravity: mgsin(theta)
Perpendicular gravity: mgcos(theta)
Normal force (in the perpendicular direction): mgcos(theta)

I then found the necessary inward force to maintain a circle:
Inward force: (v^2)/R * m = 400/125 * m = 3.2m
Separated into parallel and perpendicular components:
Parallel: 3.2mcos(theta)
Perpendicular: 3.2msin(theta)

If the car is not sliding and there is no friction, then the parallel gravity force is equal to the parallel component of this necessary inward force: 3.2mcos(theta)=mgsin(theta) this leads to theta=18 degrees

Now if a car is moving 30m/s, then the new necessary inward force is 900/125 * m = 7.2m
The components of this are: parallel: 7.2mcos18 , Perpendicular: 7.2 sin18
The friction can be denoted as f=u*normal force=u*mgcos18

If the car is not sliding then 7.2mcos18= mgsin18+ umgcos18

This solves out to u = .41

The answer is supposed to be .33

Where did I go wrong?

Second problem:

## Homework Statement

The 4.00- block in the figure is attached to a vertical rod by means of two strings. When the system rotates about the axis of the rod, the strings are extended as shown in the diagram and the tension in the upper string is 79.0 .

A) What is the tension in the lower cord?
B)How many revolutions per minute does the system make?
C)Find the number of revolutions per minute at which the lower cord just goes slack.

## Homework Equations

F=ma
Centripetal acceleration = (v^2)/R

## The Attempt at a Solution

(1^2) + (L^2) = (1.25^2) , solving this gives me that the mass is .75 m from the rod
Arctan(1/.72) = theta , theta = 53.13

Necessary inward force: (v^2)/R * m

For upper string: ((v^2)/.75) *m = 79cos 53.13 this leads to m*(v^2) =35.55

That was all I could figure out so far.

Any help would be greatly appreciated. I have been working on this problems for a day and a half, and I just can't seem to solve them. Thanks in advance.

thebrandon said:

## Homework Statement

A curve with a 125 radius on a level road is banked at the correct angle for a speed of 20. If an automobile rounds this curve at 30 , what is the minimum coefficient of static friction between tires and road needed to prevent skidding?

Normal force (in the perpendicular direction): mgcos(theta)
Don't assume this. Instead, treat the normal force as an unknown.

thebrandon said:
(1^2) + (L^2) = (1.25^2) , solving this gives me that the mass is .75 m from the rod
Arctan(1/.72) = theta , theta = 53.13

Necessary inward force: (v^2)/R * m
OK. That's the net inward force on the mass.

For upper string: ((v^2)/.75) *m = 79cos 53.13 this leads to m*(v^2) =35.55
The tension from the upper string is not the only force acting on the mass.

Doc Al said:
Don't assume this. Instead, treat the normal force as an unknown.

Why can I not assume this? It was my understanding that the normal force is equal to the total force acting on the surface. Isn't the perpendicular component of gravity the only force acting on the surface?

Also, if I accept that I can't assume this, then I would be left with the following:
7.2mcos18= mgsin18+ u(normal force), which simplifies to U=3.802m/(normal force)
I am not sure where I would go from there.

Can you please clarify as to how I am supposed to find the normal force?

Thanks

thebrandon said:
Why can I not assume this? It was my understanding that the normal force is equal to the total force acting on the surface. Isn't the perpendicular component of gravity the only force acting on the surface?
Sure, the normal force is the force acting on the surface. But no, it's not equal to the component of gravity. (It would if there were no acceleration.)

Consider vertical and horizontal forces acting on the car and apply Newton's 2nd law like usual.

Doc Al said:
Sure, the normal force is the force acting on the surface. But no, it's not equal to the component of gravity. (It would if there were no acceleration.)

Consider vertical and horizontal forces acting on the car and apply Newton's 2nd law like usual.

Ok. Thank you so much for the quick responses. You were extremely helpful. I believe I understand my error now. I was not considering the perpendicular component of acceleration necessary to maintain a circle. If I add that into account that gives me the following:
(normal force) = mgcos18 + 7.2msin18
Substituting in this value of the normal force into the equation i calculated for u gives me:
u=3.802/(9.8cos18 + 7.2msin18)= 0.33
This is the correct answer. Did I find it by chance, or have I done everything correctly now?

As per my second problem, I will now read your earlier suggestion on that, and attempt a solution. Thank you so much.

thebrandon said:
Ok. Thank you so much for the quick responses. You were extremely helpful. I believe I understand my error now. I was not considering the perpendicular component of acceleration necessary to maintain a circle. If I add that into account that gives me the following:
(normal force) = mgcos18 + 7.2msin18
Substituting in this value of the normal force into the equation i calculated for u gives me:
u=3.802/(9.8cos18 + 7.2msin18)= 0.33
This is the correct answer. Did I find it by chance, or have I done everything correctly now?
That's a perfectly fine way to solve the problem. Looks good to me.

Ok. For the second problem I have done the following:

(1^2) + (L^2) = (1.25^2) , solving this gives me that the mass is .75 m from the rod
Arctan(1/.72) = theta , theta = 53.13

Net horizontal force: (v^2)/R * m

Adding the horizontal components of the tension of both ropes and setting that equal to the net horizontal force gives me:
79cos53.13 + (Tension of second rope)cos53.13 = m(v^2)/.75
this simplifies to:
(tension of second rope)=((V^2)-8.89)/0.11

I can't figure out what the velocity is though, so I am not able to finish this solution.

Any suggestions?

thebrandon said:
Any suggestions?
Analyze the vertical forces acting on the mass.

## 1. What is circular motion?

Circular motion is the motion of an object along a circular path. It is characterized by a constant speed and a continuous change in direction.

## 2. How is friction involved in circular motion?

Friction is involved in circular motion because it acts as a force that resists the motion of an object along a curved path. This frictional force allows the object to maintain its circular motion by continually changing its direction.

## 3. How does tension affect circular motion?

Tension is a force that is often involved in circular motion, especially when an object is attached to a string or rope. Tension helps to keep the object moving along the circular path by providing the necessary centripetal force.

## 4. What is the difference between centripetal and centrifugal force?

Centripetal force is the force that acts towards the center of a circular path and keeps an object moving along that path. Centrifugal force, on the other hand, is a fictitious force that appears to act outwards from the center of the circular path, but in reality, it is just the reaction force to the centripetal force.

## 5. What are some real-life examples of circular motion?

Some common examples of circular motion include a car turning around a bend, a ball spinning around in a circle, and a planet orbiting around the sun. Other examples can include amusement park rides, a spinning top, and a swinging pendulum.

Replies
6
Views
1K
Replies
2
Views
1K
Replies
11
Views
2K
Replies
6
Views
2K
Replies
2
Views
1K
Replies
14
Views
3K
Replies
8
Views
2K
Replies
2
Views
2K
Replies
7
Views
2K
Replies
1
Views
1K