Circular Motion Dynamics. Car traveling in vertical circle

[Unix]

Homework Statement

A small car with mass .800 kg travels at a constant speed of 12m/s on the inside of a track that is a vertical circle with radius 5.0m. If the normal force exerted by the track on the car when it is at the top of the track is 6.00N, what is the normal force at the bottom of the track?

Homework Equations

The Attempt at a Solution

At the top of the track the FBD has the normal force pointing downwards, the weight is pointing downwards.

At the top of the track the acceleration is pointing downward so the equation using Newtons laws is

N+mg=ma

At the bottom of the circular track the normal is force is pointing upwards and the weight is pointing downwards and the acceleration is pointing upwards

N-mg=ma

The normal force at the top of the track was given. So I used that to solve for the acceleration

6+(.800)(9.8)=(.800)a

after solving for "a" I plugged the acceleration into the second equation to solve for the normal force at the bottom. The answer was 27.1 N. This is exactly what my solutions manual has,

but my question is: If we are in circular motion, there is a given velocity of 12m/s and a radius of 5m/s. Why can't we use a=(V^2)/R to find the acceleration ?If i use v^2/r I get a different answer for my acceleration but I don't understand why

 

[Doc Al]

but my question is: If we are in circular motion, there is a given velocity of 12m/s and a radius of 5m/s. Why can't we use a=(V^2)/R to find the acceleration ?If i use v^2/r I get a different answer for my acceleration but I don't understand why

You most certainly should be able to use a = V^2/R. I think the problem is flawed--the given data are inconsistent.

 

[Unix]

You most certainly should be able to use a = V^2/R. I think the problem is flawed--the given data are inconsistent.

That's what I was thinking as well. There have been several "loop the loop" problems that are similar to this one where roller coasters are moving at constant velocities around circular tracks, and I've been able to find accelerations by using (V^2)/R.

NOTE:

I just realized something. The actual problem itself does not explicitly state that the velocity is 12m/s. However there is a diagram of the problem in the textbook and THAT is where it shows a velocity vector 12m/s

So the problem statement really is

"A small car with mass .800 kg travels at a constant speed on the inside of a track that is a vertical circle with radius 5.0m. If the normal force exerted by the track on the car when it is at the top of the track is 6.00N, what is the normal force at the bottom of the track?"

Since the problem doesn't explicitly state that the velocity is 12m/s. Is it correct to assume that the value of the velocity is irrelevant if we are given the normal force that is exerted at that period of time? (because we can find the acceleration directly using Newtons laws).

 

[haruspex]

Is it correct to assume that the value of the velocity is irrelevant if we are given the normal force that is exerted at that period of time? (because we can find the acceleration directly using Newtons laws).

It's not irrelevant, exactly, but you don't need to be told it. There is enough other information provided.

 

[Doc Al]

Since the problem doesn't explicitly state that the velocity is 12m/s. Is it correct to assume that the value of the velocity is irrelevant if we are given the normal force that is exerted at that period of time? (because we can find the acceleration directly using Newtons laws).

I would not say that the speed is irrelevant, just that you do not need to be told it explicitly (nor do you need to calculate it to solve the problem). You have all the information needed to solve the problem without it. (I am agreeing with haruspex.)

If that diagram is meant to go with this particular problem, then it is mislabeled with an incorrect velocity. I would stick to the data provided in the problem statement in solving the problem (as you did).

 

[md1327]

Unix, I think you misprinted the answer. I got 21.7 N for the force at the bottom.