Speed & Force in Toy Car on Frictionless Ramp

  • Thread starter StephenPrivitera
  • Start date
  • Tags
    Energy
In summary, the conversation discusses a problem involving a toy car rolling down a ramp and making a circular loop at the bottom. The speed of the car at the bottom and top of the loop, as well as the force exerted by the track at the top of the loop, are calculated using energy conservation laws. The minimum value of H for the car to successfully complete the loop is found to be greater than 5/2 times the radius of the loop. Different methods for solving the problem are mentioned, such as variational principle and Lagrange multiplier method, but the question is to be solved using energy conservation laws.
  • #1
StephenPrivitera
363
0
A toy car of mass M rolls down a frictionless ramp of height H > 2R and makes a circular loop of radius R at the bottom. (a) What is the car's speed at the bottom of the loop (b) at the top of the loop? (c) What is the force exerted by the track at the top of the loop? (d) What is the minimum value of H such that the car goes around the loop without falling off due to gravity?
The first two parts are simple.
a) Ei = MgH = Ef = 1/2Mv^2 solve for v
b) Ef = 1/2Mv^2 + 2MgR = Ei solve for v
But then...
c) The best I can do for this is figure that the track should exert a force downward on the car at the top of the track. So the forces acting on the car are Fnet = mg + Ft = ma, where Ft is the force exerted by the track. If I could find a, I could find Ft. But I can't do either.
I haven't even tried part d yet.
Help, anyone?
 
Physics news on Phys.org
  • #2
Hi!
Minimum H can be calculated by:
MgH=1/2Mv^2+2MgR (1).
Forces which is exerted on the car are -Mg, centrifugal force=MR&omega^2 and the force, T, which the surface of the track exerts on the car. Here, R&omega=v(speed of the car in the loop). So,
0<T=MR&omega^2-Mg=Mv^2/R-Mg (2).
Equation (2) means the car is attached on the surface of the track, i.e., the car exerts the force on the surface of the track, then .an opposite force is exerted on the car.
Simultaneously solving the equation (1) and (2), we have:
H>5/2R.
If there is any mistake, please correct, anyone!
Please refer to the book of classical mechanics to be sure that the expression for the centrifugal force here is correct.
 
  • #3
Originally posted by shchr
Forces which is exerted on the car are -Mg, centrifugal force=MR&omega^2 and the force, T, which the surface of the track exerts on the car. Here, R&omega=v(speed of the car in the loop). So,
0<T=MR&omega^2-Mg=Mv^2/R-Mg (2).
I haven't learned about the centrifugal force. The chapter is on energy conservation, so I assume the problem is supposed to be solved using those laws.
 
  • #4
I do not know how the problem is solved only by energy conservation law. Did you learn about variational principle? If so, you must know Lagrange multiplier method. Using this technique, you can get the same result.
 
  • #5
Nope. I've never heard of any of these. The question comes from volume 1, chapter 7, question# 36 in "Physics for Scientists and Engineers" by Fishbane et al. I guess I'll try not to let it keep me awake at night, but it would be nice to see a solution.
 

1. What is the relationship between speed and force in a toy car on a frictionless ramp?

The relationship between speed and force in a toy car on a frictionless ramp is that the speed of the car is directly proportional to the force applied. This means that the greater the force applied, the faster the car will move.

2. How does friction affect the speed and force of a toy car on a frictionless ramp?

Since the ramp is frictionless, there is no resistance to slow down the car. Therefore, friction does not affect the speed and force of the car on the ramp. However, if there is friction on the surface the car is rolling on, it will slow down the car and decrease the force needed to move it.

3. Can the speed of a toy car on a frictionless ramp reach a maximum limit?

Yes, the speed of a toy car on a frictionless ramp can reach a maximum limit. This is due to the conservation of energy, where the potential energy of the car at the top of the ramp is converted into kinetic energy as the car moves down the ramp. Once all of the potential energy is converted, the car will reach its maximum speed.

4. How does the angle of the ramp affect the speed and force of the toy car?

The steeper the angle of the ramp, the greater the force needed to move the car. This is because the steeper angle will increase the gravitational force acting on the car, accelerating it more quickly. However, the speed of the car will also increase as it moves down the steeper ramp.

5. Can the mass of the toy car affect its speed on a frictionless ramp?

No, the mass of the toy car does not affect its speed on a frictionless ramp. This is because, in the absence of friction, the mass of an object does not affect its acceleration. Therefore, a heavier car and a lighter car would reach the bottom of the ramp at the same speed, as long as the force applied is the same for both cars.

Similar threads

Replies
7
Views
1K
  • Mechanics
Replies
3
Views
240
  • Introductory Physics Homework Help
Replies
13
Views
861
  • Introductory Physics Homework Help
Replies
2
Views
528
  • Introductory Physics Homework Help
Replies
18
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
955
  • Introductory Physics Homework Help
Replies
12
Views
10K
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Other Physics Topics
Replies
1
Views
26K
Back
Top