Physics Question pt. 1

Question

I'm really struggling with this problem. If anyone could help me out with these it would be greatly appreciated:

1.) An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is 5.2 kN, and the radius of the circle is 12 m. At the top of the circle, (a) what is the force FB on the car from the boom (using the minus sign for downward direction) if the car's speed is v = 5.7 m/s? (b) What is FB if v = 17 m/s?

 

Thanks

Answer 1

Lisa,

Wyzant isn't intended to answer your physics problem sets -- 5 problems at a time is spamming. It's to get you some spot help, past difficulties of conceptualizing what you have to do.

I'll help you with this one, though, and perhaps you can apply the type of thinking to the other problems (and retract them if you do solve them)?

View the ride from the side. The cars are moving in a circle. Circular motion has some specific equations of movement and force associated with it, namely, a = v²/r (among others!). In this equation a = the acceleration of the object (here, the car+riders), v = the velocity of the object (here, 5.7 m s^-1 ), and r = the radius of the circle of motion (here, 12 m). You are given a little additional information, namely the weight of the object in kN. Since this is already a force, you know that it is the mass multiplied by a(gravity) == 9.8 m s^-2 . Thus, m = (5.2/9.8)*1000 kg

Now you are ready to consider the car, in its path of motion. You know that the force on it, at the top of the circle, is whatever is sufficient to keep it moving in the circular path. And you should know, by this point in your course, that the acceleration of an object in a circular path (at a uniform speed) is always exactly directed towards the center of the circle of rotation. You know one piece of this force: the static gravity force = -5.2kN (using the conventional sign for direction of forces). That plus the force supplied by the arm, must total F = ma = -(5.2/9.8)*1000 kg * (5.7)² (m s^-1)² / 12 m -- that is, the total force providing the acceleration required, right at the top of the circle. You should be able to handle the math from there?