# B Rotating Sphere: Conceptual Question

#### Shivansh

Summary
Finding angular acceleration of rotating Sphere. 3 methods 3 different answers

As shown in figure there's a homogeneous solid sphere. It is rotating about axis which is passing through point P directed perpendicular to the plane of paper. (In short like a pendulum).

I'm neglecting gravity and assuming a force F which is directed perpendicular to the string. (The string, if extended will pass though the centre of the sphere, also note that the force will be continuously changing its direction so as to always maintain 90° angle).

It's clear that the only external forces acting on the sphere are Tension and F (both perpendicular). The centre of mass moves in a circle (due to constraint by string).

Suppose I have to calculate angular acceleration of the sphere. There are three ways each yielding a different value. How is it possible?

Method 1: Centre of mass moves in circle about point P with tangential force F and radial force T. The angular acceleration of COM is simply Tangential Force (F) divided by product of distance OP and mass of body. (Only external forcees act on centre of mass, here external forces are F and Tension T)

Method 2: The axis passing through O is stationary. The torque about this axis is obvious. The moment of inertia of body about given axis is known. Using this the angular acceleration is found to be lesser than that found using method 1. (Please see uploaded picture at the end of question to see the exactly about the methods)

Methods

EDIT: Why isn't someone replying? Is there something wrong with the question fell free to point out the mistake.

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#### Andrew Mason

Homework Helper
The angular acceleration depends upon the torque, which depends not only on the magnitude and direction of the force but also the distance from the pivot of the point of application.

AM

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#### Shivansh

The angular acceleration depends upon the torque, which depends not only on the magnitude and direction of the force but also the distance from the pivot of the point of application.

AM
But there is a torque about point P (located where the thread is attached with the wall).

#### Andrew Mason

Homework Helper
But there is a torque about point P (located where the thread is attached with the wall).
If you apply the force F to the centre of the sphere, the mass accelerates with acceleration a = F/m. So the angular acceleration is F/ml.

If you apply the force other than to the centre of mass of the sphere, you apply an additional torque that causes the sphere to rotate about its centre of mass as well as causing the centre of mass to accelerlate. Then you have to take into account the moment of inertia of the sphere.

Welcome to PF by the way!

AM

#### A.T.

...the string, if extended will pass though the centre of the sphere...
This is not correct. The sphere has a changing angular velocity around its own CoM, so the string force must provide some torque around that CoM.

#### erbahar

A.T is right. Your string or rod must supply a tangential component. That complicates the situation regarding com acceleration.

#### Andrew Mason

Homework Helper
I think we can assume that the string's mass is negligible compared to the mass of the sphere.

In method 1, if the string is attached to the surface of the sphere, A.T. is correct that the string will apply a torque to the sphere that will cause the sphere to rotate about its centre of mass. That torque will be have a component that opposes the applied torque. If the string is extended and attached to the centre of the sphere, then the string does not apply a torque as the sphere moves (like a yo-yo except that it is spherical).

AM

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#### Andrew Mason

Homework Helper
Just to follow up, if the string is attached to the surface of the sphere so that the sphere is rotating as well, the string + sphere acts as a rigid body. This takes into account the rotational torque on the sphere. The moment of inertia determined using the parallel axis theorem is $\frac{2}{5}mr^2 + mL^2$. Assuming the force is applied to the centre of mass of the sphere and assuming that the mass of the string is negligible, then:

$I\alpha = \tau = FL$
$\alpha = FL/(\frac{2}{5}mr^2 + mL^2)$

AM

#### A.T.

...if the string is attached to the surface of the sphere so that the sphere is rotating as well, the string + sphere acts as a rigid body.
An idealized string only transmits forces parallel to the string, and no lateral components. If string + sphere would act as a rigid body, the string could not exert a varying torque around the CoM. I think what you mean is a sphere on a mass-less rod, not a string.

#### Andrew Mason

Homework Helper
An idealized string only transmits forces parallel to the string, and no lateral components. If string + sphere would act as a rigid body, the string could not exert a varying torque around the CoM. I think what you mean is a sphere on a mass-less rod, not a string.
Quite correct. That is good to point out.

But if the result is that the sphere rotates so that the contact point between the string and the sphere stays in approximately the same position relative to a line from P to O, the system behaves very much like a rigid body. This means that the string is not on the line OP but angled a bit forward of it. As a result the applied force and the tension in the string provide a torque to the sphere. That torque causes the sphere to experience angular acceleration about its COM that is identical to the angular acceleration of the sphere COM about point P. As the angular speed increases, the tension in the string increases and the angle of the string to OP decreases.

AM

#### Shivansh

Quite correct. That is good to point out.

But if the result is that the sphere rotates so that the contact point between the string and the sphere stays in approximately the same position relative to a line from P to O, the system behaves very much like a rigid body. This means that the string is not on the line OP but angled a bit forward of it. As a result the applied force and the tension in the string provide a torque to the sphere. That torque causes the sphere to experience angular acceleration about its COM that is identical to the angular acceleration of the sphere COM about point P. As the angular speed increases, the tension in the string increases and the angle of the string to OP decreases.

AM
You are correct. I was confused on *exactly* how the string will apply torque. I now understand that the string won't pass through COM of sphere when extended and will be making some angle with the contact point so as to provide net torque. So my methods 1 and 3 are wrong and second one is right. Thanks for your reply!

"Rotating Sphere: Conceptual Question"

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