Speed & Force in Toy Car on Frictionless Ramp

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The discussion revolves around a toy car rolling down a frictionless ramp and its motion through a circular loop. For part (a), the car's speed at the bottom of the loop is derived from energy conservation, resulting in the equation MgH = 1/2Mv^2. In part (b), the speed at the top of the loop is calculated using energy conservation again, leading to the expression involving potential energy at height 2R. The force exerted by the track at the top of the loop is analyzed in part (c), where the net forces are considered, and it is concluded that the minimum height H for the car to complete the loop without falling off is H > 5/2R. The conversation highlights the importance of energy conservation principles while also touching on the concept of centrifugal force, though some participants express confusion about these concepts.
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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?
 
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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.
 
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.
 
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.
 
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.
 
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