6
$\begingroup$

Given this question:

A small ball of mass $m$ and radius $r$ rolls without slipping on the inside surface of a fixed hemispherical bowl of radius $R>r$. What is the frequency of small oscillations?

The standard solution is to write Newton's second law for the ball and then take the centre of mass of the ball to be the pivot and write

$$\tau = I \alpha.$$

Only the frictional force contributes to the torque in this case. From Newton's second law, I can express the frictional force in terms of the gravitational force and therefore the frictional force can be eliminated in the equation for torque. I then make small angle approximation and get the equation to be of the form

$$k\theta=-I\ddot{\theta}$$

from which I can find the frequency.

Another approach uses the point of contact of the ball with the sphere as the pivot. It has the advantage that the frictional force adds no torque. Both approaches give the same result.

My question is since both pivots that we have chosen are accelerating, why are not fictitious forces considered? In the first place, can the pivots that we choose when writing

$$\tau = I \alpha$$

be accelerating?

$\endgroup$
1
  • $\begingroup$ Forces through the c.m. play no role in the angular equations of motion. $\endgroup$ Commented Feb 27, 2014 at 18:20

2 Answers 2

1
$\begingroup$

In fact, you are right in that you should consider the fictitious force in accelerated reference frame. As such, using the point of contact between the mass and the bowl is actually not ideal. In the center of mass reference frame, notice if you consider a fictitious force for the accelerated reference frame, it is applied at the center of mass, and thus it does no torque. This is better.

However, the center of mass and the point of contact between the ball and the bowl are actually not the points of reference that I would use. Instead, I would use the center of curvature of the bowl, because this point of reference is not accelerating and still has the benefit of excluding the normal force from the torque equation.

tl;dr: You can only ignore fictitious force in the center of mass reference frame, but otherwise you are right in needing to include it.

$\endgroup$
-1
$\begingroup$

My question is since both pivots that we have chosen are accelerating, why are not fictitious forces considered?

Pivot is accelerating if it is a geometric point defined as the point of contact. Pivot is not accelerating if it is the material point of the small ball; it stands still and has zero acceleration. The description resulting in the equation of motion is done in the frame of the bowl, which is considered as inertial frame, hence no inertial forces.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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