Calculating Forces on a Motorcycle Riding in a Transparent Sphere

  • Thread starter Thread starter solarcat
  • Start date Start date
  • Tags Tags
    Motorcycle
Click For Summary
The discussion revolves around calculating the forces acting on a motorcycle inside a transparent sphere while moving in a horizontal circle. Key points include determining the minimum coefficient of static friction required to prevent tire slippage and understanding the forces at play, such as weight, normal force, and friction. The normal force is linked to the centripetal force equation, mv²/r, while friction acts to oppose potential slipping. The conversation emphasizes the importance of free body diagrams to visualize these forces and clarify the relationship between them. Ultimately, the analysis focuses on maintaining constant velocity and the implications of tire slippage on motion.
solarcat
Messages
22
Reaction score
3

Homework Statement


A person is riding a motorcycle in a hollow, transparent plastic sphere in a horizontal circle. The radius is 13 m, the mass of the motorcycle is 40 kg, and the mass of the person is 70 kg. The speed is 15 m/s.
a) What is the minimum coefficient of static friction for the tires to not slip?
b) What total force does the motorcycle exert on the rider?

Homework Equations

The Attempt at a Solution


First I thought there would be a force of weight pulling the motorcycle down, a normal force of the sphere on the motorcycle, and a force of friction perpendicular to the normal force/opposite the direction of motion. In this case, wouldn't the normal force be equal to mv^2/r? But then how does this tell you when the tires will slip?

Then I was looking at some diagrams of cars traveling in circles, and the diagram showed the friction force going to the center/perpendicular to the direction of motion. If this is the case, then wouldn't mv^2/r be equal to the sum of the normal force and the friction force? But then how do you know what the normal force even is?
 
Physics news on Phys.org
I assume there is also an assumption that this occurs along the equator of the sphere?

The friction force along the direction of motion is only necessary of you want to accelerate. This is a motion with constant velocity. I suggest you draw a free body diagram and compare with the total force necessary to keep moving on the horizontal circle.
 
So I think the free body diagram would be something like this...
And you would set up the equations
N( 1 +us) = mv^2/r
N(1+us) = (110 kg) (15 m/s)^2 / (13 m) = 1904 N
But I have two unknowns so I'm not sure how to solve it. Also, the second part seems to imply that I should find the normal force which means it shouldn't be needed for the first part.
 

Attachments

  • Screen Shot 2019-03-01 at 2.37.53 PM.png
    Screen Shot 2019-03-01 at 2.37.53 PM.png
    9.6 KB · Views: 378
solarcat said:
perpendicular to the normal force/opposite the direction of motion.
Friction opposes relative motion of surfaces in contact. In rolling contact, you have to think in which direction slipping would occur without friction; the frictional force acts to oppose that.
solarcat said:
wouldn't the normal force be equal to mv^2/r?
mv2/r is the centripetal acceleration. Since the speed is constant that is the total acceleration and results from the net force: ΣF=ma. You listed all the forces.
 
  • Like
Likes solarcat
haruspex said:
Friction opposes relative motion of surfaces in contact. In rolling contact, you have to think in which direction slipping would occur without friction; the frictional force acts to oppose that.

I think you're asking to determine the direction of slip first so you can then determine the direction of friction. But if the wheels do slip, then the motorcycle won't move, so then doesn't mean there's no acceleration and therefore no force?
 
solarcat said:
But if the wheels do slip, then the motorcycle won't move
This is incorrect. Again, draw the free body diagram.

Note that there is no force in the direction of travel. That would lead to an increase in speed.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 2 ·
Replies
2
Views
796
  • · Replies 97 ·
4
Replies
97
Views
6K
  • · Replies 21 ·
Replies
21
Views
1K
Replies
23
Views
13K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
6
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 7 ·
Replies
7
Views
12K
  • · Replies 3 ·
Replies
3
Views
2K