Acceleration due to Centripetal force

I'm thinking about how this sphere would move down the ramp and through the loop at the bottom of the ramp. I'm mostly interested in whats going on at the very top point of the loop. If I were to draw a free-body diagram of the sphere at that point, what forces would be present? I'm guessing that there would be weight directed downwards, and then the centripetal force pointing upwards. Is this correct? The centripetal force appearing on a FBD of the sphere at this point would be a force pair? As in, the centripetal force is provided by the loop surface on the sphere directed towards the center of the loop, so with equal and opposite force pairs, the sphere would exert the same but opposite force on the loop? I'm just trying to straighten out in my head where the normal force from the surface is. I always thought that if there is contact between a surface and object, there would be a normal force. Where is it in this case?

If i were trying to solve for something like the minimum height needed for the sphere to pass through the loop, I'm solving for when the centripetal force equals the weight...but again, where is the normal force during all of this?

rcgldr
Homework Helper
I'm guessing that there would be weight directed downwards, and then the centripetal force pointing upwards.
Centripetal force would also be downwards.

Remember that with free body diagrams, you only incorporate those forces acting ON the object. You mention the force the sphere exerts on the ramp; this upward force would not be present on a FBD for the SPHERE. only the force the ramp exerts ON the sphere which is downward and that is your centripetal force.

Doc Al
Mentor
If I were to draw a free-body diagram of the sphere at that point, what forces would be present? I'm guessing that there would be weight directed downwards, and then the centripetal force pointing upwards. Is this correct?
No, for two reasons:
(1) The 'centripetal force' acts towards the center and thus downward.
(2) The 'centripetal force' is just the name given to the net radial force associated with circular motion; it's not something that should appear on a free body diagram.

The actual forces acting on the sphere at the top of the loop are the weight and the normal force from the surface, both of which act downward. That's all that should appear on a FBD.
The centripetal force appearing on a FBD of the sphere at this point would be a force pair? As in, the centripetal force is provided by the loop surface on the sphere directed towards the center of the loop, so with equal and opposite force pairs, the sphere would exert the same but opposite force on the loop?
The sphere and the surface exert equal and opposite forces on each other--the normal force. (Not the 'centripetal force'.)
I'm just trying to straighten out in my head where the normal force from the surface is. I always thought that if there is contact between a surface and object, there would be a normal force. Where is it in this case?
Right where you think it is.
If i were trying to solve for something like the minimum height needed for the sphere to pass through the loop, I'm solving for when the centripetal force equals the weight...but again, where is the normal force during all of this?
What you should be doing is applying Newton's 2nd law. The condition for the minimum speed to maintain contact at the top of the loop is that the normal force should just be zero.
ƩF = ma
ƩF = mv2/r

wow, I was very mixed up it looks like, but this helps a ton. thank you guys.

rcgldr
Homework Helper
If i were trying to solve for something like the minimum height needed for the sphere to pass through the loop.
If the sphere is rolling (and never slides) instead of sliding on a frictionless surface, you'll need a greater initial height since some of the gravitational potential energy will be converted into angular kinetic energy.

Ken G
Gold Member
Indeed, it does sound like the sphere should be rolling, or else there is no need to say that it has a spherical shape (which brings in its moment of inertia). What's more, if the sphere is indeed rolling, then there is one more force in the FBD-- the static friction force. However, it will turn out this force is zero at the top of the loop, because the rate of spin of the sphere will not be changing there. So it as rcglder says-- a rolling sphere simply needs to have a kinetic energy of rotation, and it has to be consistent with its kinetic energy of motion along the track, such that the point of contact is always instantaneously stationary. That constraint connects the spin rate to the motion, and the other constraint is that the total kinetic energy equals the loss of gravitational potential energy. With those two constraints, you won't need a FBD at all.

Doc Al
Mentor
With those two constraints, you won't need a FBD at all.
Regardless of whether the sphere slides or rolls you'll need a FBD just the same.

Ken G
Gold Member
Regardless of whether the sphere slides or rolls you'll need a FBD just the same.
Yes, since there is an issue whether or not the sphere will reach the top. At any point, you know the total kinetic energy by looking at the gravitational potential difference. The rolling speed detemines both the translational kinetic energy and the rotational kinetic energy of the sphere, and they must add up to the total. This gives you the speed at every point, you just don't know if it reaches the top because you don't know if you are implicitly assuming the normal force can be negative. So that part does require a FBD to some extent, and would be true for sliding or rolling, but the answer is different if it is rolling. Even more to the point, if it is rolling, a normal FBD must include the static frictional force, but you won't need that if you are doing it with conservation of energy. All you need to know is the normal force, which must add with the appropriate component of gravity to give the centripetal force needed to follow the loop, given the speed that is found from conservation of energy. So all you need is the component of the FBD that points perpendicular to the track, but you're right, you will need that if you want the minimum height that reaches the top.

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