(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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?

2. Relevant equations

3. 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

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# Homework Help: Circular Motion Dynamics. Car traveling in vertical circle

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