You are here:

# Physics/Physics, mechanics

Question
I am having difficulty understanding this problem. Any help would be gretaly appreciated.

An amusement ride consists of a car moving in a vertical circle on the end of a rigid boom. The radius of the circle is 10 m. The combined weight of the car and riders is 5.0 kN. At the top of the circle the car has a speed of 5.0 m/s which is not changing at that instant. What is the force of the boom on the car at the top of the circle?

This problem is a Newton's 2nd Law problem.
The first step is to develop the appropriate free body diagram.
In this case there are 2 relevant forces: the force of gravity Fg downward and the force of the boom Fb direction unknown - let's assume that it is down. If it turns out to be up the answer will come out negative.
The second step will be to determine the direction of the acceleration. In this case the acceleration is centripetal and is, therefore, directed towards the center of the circular path which at the top of the ride is directed downward.
The third step is to apply Newton's 2nd Law: SF=m*a where SF is the sum of the forces in the direction of th acceleration.
In this case this becomes:
SF=m*a
Fg+Fb=m*a=m*v^2/r where a=v^2/r is th centripetal acceleration.
Therefore, the force of the boom will be (Assuming g=10):
Fb=m*v^2/r-Fg=5000/10*5^2/10-5000=1250-5000=-3750N
So the force of the boom must be 3750N up!

Physics