A level physics circular motion

I would just like to quickly clarify something in circular motion because I am getting different answers in past papers. In vertical circular motion, at the top of the loop does the: 1. object's weight provide all the centripetal force required with no contact force needed (contact force is zero) OR 2. object's weight provide SOME of the required centripetal force but some of the normal contact force is still needed. Thank you for your help.

Reply 1

Mr Naga Physics

Weight is constant and downwards ( W = mg , mass doesn't change, g doesn't change).

Normal contact force can change. For exampel over a bridge, the normal contact force is upwards. It is smaller than weight so that the resultant force is towards the center of the circle. (N-mg = mv^2/r). As speed increases, N decreases until it is zero at which point you lose contact with the surface.

If you're talking about a loop-the-loop type of vertical motion then normal contact is downwards , N+mg = mv^2/r and the slower you go the smaller N becomes and at N = 0, the rollercoaster might fall off the track ( too slow).

You need to learn to draw the freebody diagram with the arrows with correct relative lengths to show which is bigger mg or N.

Reply 2

Eimmanuel

Study Forum Helper

Original post

by JibberJam

It depends on what the question is asking.

I assume you are referring to the vertical circular motion shown in the diagram above.

If the question is asking what is the minimum speed that the vehicle needs to have at the top of the circular track to undergo a complete circular motion, then the contact force at the top is zero and the weight provides the centripetal force for the vehicle to complete the circular motion.

If the question is asking what is the contact force at the top when the vehicle is moving at a speed greater than $\sqrt{rg}$ , then the contact force and weight provide for the centripetal force.