0
$\begingroup$

I ran across a conceptual problem while solving this situation:

A marble is sliding down from the top of a larger sphere, at what angle from the vertical will it leave the sphere?

I solved the problem by using the fact that the marble will leave the sphere when the radial component of the gravitational force is equal to the centripetal force the marble is experiencing. My problem is this: why is there centripetal acceleration at all?

I understand that the marble is moving in a circular path while on the sphere, but why doesn't all of the gravitational force in the radial direction go into the normal force? I know this would mean that the only force acting on the marble would be a tangential force so it's not correct, but why?

I guess my problem with it is that if we were to pick up the marble and place it an inch further on the sphere, there wouldn't be a centripetal force, all of the radial force would be normal; so when does it become centripetal and why? Also, isn't circular velocity increasing as the ball falls, so wouldn't that change the centripetal acceleration?

Sorry if this question is a bit trivial, but I'm just confused. Thank you in advance for your help.

$\endgroup$
  • $\begingroup$ It sounds like you "solved" the problem without understanding what you were doing. $\endgroup$ – sammy gerbil Sep 3 '16 at 19:45
  • $\begingroup$ That's true. I knew what had to happen, but I didn't know why. Which is why I came here to understand why there had to be centripetal acceleration. $\endgroup$ – B. Sandoval Sep 3 '16 at 19:49
2
$\begingroup$

It is better that the term "centripetal force" is not used. In this example any net force along the line joining the two centres of the spheres (which have called the radial direction) will cause a radial acceleration.
The small sphere wants to move downwards due to the gravitational attraction of the Earth.
It cannot do so because the large sphere exerts a force on it which is called the normal reaction.
So the net force on the small sphere is the sum of those two forces.
That net force can be resolved into two components to make the analysis of the motion easier.
In this example the chosen components are radial and tangential at the point of contact between the small and large sphere.
When the spheres are in contact the trajectory of the small sphere is an arc of a circle whose radius is that of the large sphere.
The radial component of the net force causes a radial (centripetal) acceleration.

If the radial component of the net force was zero, normal reaction equals radial component of gravitational force, there would be no radial acceleration, only a tangential acceleration and the small sphere would lose contact with the large sphere.

If by moving an inch you mean that the small sphere starts from rest at a point which is lower than the top of the large sphere then at the instant of release the radial and normal reaction forces are equal and opposite.
There is no centripetal acceleration as the small sphere is not moving.
However as soon as the small sphere starts to move due to the net tangential force on it the normal force becomes smaller than the radial force and net inward radial force produces a radial acceleration.
If that net radial force did not exist the motion of the small sphere would be along a tangent to the large sphere due to the tangential force and the small sphere would lose contact with the large sphere.

$\endgroup$
  • $\begingroup$ Thank you for the very thorough answer! I just have one question, does the tangential acceleration affect the centripetal acceleration? As I understand it $a=\frac {v^2}{r}$ , so if circular velocity is increasing, won't the centripetal acceleration? $\endgroup$ – B. Sandoval Sep 3 '16 at 6:37
  • $\begingroup$ You are looking at what happens at an instant of tine when the tangential speed is $v$. At that time the acceleration of the sphere has two complements: the tangential and the radial $\frac{v^2}{r}$. At the next instant the speed has increased and so will the radial acceleration. $\endgroup$ – Farcher Sep 3 '16 at 7:16
  • $\begingroup$ Ah, I see. That is why the normal force becomes smaller as the marble slides; because more of the gravitational force in the radial direction needs to go into increasing the centripetal acceleration. $\endgroup$ – B. Sandoval Sep 3 '16 at 7:21
  • $\begingroup$ Also note that the angle between the radial component of the gravitational attraction and the vertical increases which is reducing that radial component of the gravitational attraction. So increasing the rate at which the normal reaction is decreasing. $\endgroup$ – Farcher Sep 3 '16 at 7:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.