# Why can we treat a ball as a point mass to calculate torque?

A sphere of radius $$R$$ is supported by a rope attached to a wall as shown in the below figure. The rope makes an angle $$\theta$$ with respect to the wall. The point where the rope is attached to the ball is such that if the line of the rope is extended it crosses the horizontal line through the center of the ball at a distance $$3R/2$$ from the wall. What isthat the minimum $$μ_s$$ between wall and ball for this to be possible?

The answer claims that: The forces that do apply a torque are the force of gravity acting downward at $$C$$, so that $$\tau_g = MgR/2$$.

What I'd like to understand is why we calculate the torque of the ball due to gravity with respect to point $$O$$ by treating the ball as if it's entire mass is at point $$C$$? Is that indeed what the answer is supposing? It seems to me that the correct answer would require calculating the torque by coming up with an integral that treats the mass as evenly distributed throughout the ball.

Calculating the torque on a rigid body w.r.t to the point $$\vec 0$$ (WLOG) with gravity pointing in a constant direction $$\hat n$$ is accomplished by integrating over the rigid body, with each differential mass element $$dm$$ at position $$\vec R$$, the differential torque:

$$\tau = \int d \tau = \int \vec r \times (dm)g \hat{n} = \left( \int \vec r \, dm \right) \times g\hat{n}$$

But since the center of mass $$\vec C$$ is defined as $$\left( \int \vec r \, dm \right)/M$$, where $$M = \int dm$$ is the total mass of the rigid body,

$$\tau = \vec C \times \left( Mg \hat{n} \right)$$

The crucial assumption that reduced the torque to a cross product involving the total weight of the object $$Mg\hat{n}$$ and the center of mass (w.r.t. $$\vec 0$$) $$\vec{C}$$ is that the strength and direction of the gravitational field are constant across the rigid body. If this were not true, I could not have popped $$g$$ and $$\hat{n}$$ out of the integral above.

The torques can be calculated in respect to any point, and it generally requires taking integrals. However, there are some basic tricks to simplify life. One of them is choosing the rotation axis at the point that is located on the line of action of one of the forces, which renders the arm of this force to have zero length, resulting in zero torque. Thus, choosing point O as the rotation axis makes it unnecessary to calculate the torque due to the tension force of the rope.

Another trick is that the torque of a symmetric object (a sphere or a cylinder with axis along the rotation axis) can be calculated as if all its mass was located in the center of the sphere (axis of the cylinder). The answer by @jwimberley provides the simple proof that this is indeed the case.