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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?

Answer
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!

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I am teaching or have taught AP physics B and C [calculus based mechanics & electricity and magnetism] as well as Lab Physics for college bound students. I have a BS in Physics from the University of Pittsburgh and a Master of Arts in Teaching from same. I have been teaching physics for 34 years. I am constantly updating my skills and have a particular interest in modern physics topics.

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