# Tension and compression force

## Main Question or Discussion Point

I found a long-time ago problem in helping archive here

The problem is about attaching a mass m to a massless rod which released at initial angle $$\theta_0$$ (measured from vertically axis), try to find the critical angle where the compression force change to tension force. Using newtonian physics, the radial direction

$$T-mg\cos\theta=-ml\frac{v^{2}}{l}$$

where T is the compression/tension of the rod, v is the velocity along tangential direction. At critical angle, the compression/tension force become zero, so

$$mg\cos\theta=ml\frac{v^{2}}{l}$$

Apply the conservation of energy, one can find the critical angle is

$$\theta_c = \cos^{-1}\left(\frac{2\cos\theta_0}{3}\right)$$

which is same as the solution in the old post.

But my question is, when the mass keep falling and the angle is larger than the critical angle, will the net force pointing outward so the acceleration is not pointing toward the center? If so, the object will not do a circular motion when it pass the critical angle?

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tiny-tim
Homework Helper
Hi KFC! But my question is, when the mass keep falling and the angle is larger than the critical angle, will the net force pointing outward so the acceleration is not pointing toward the center? If so, the object will not do a circular motion when it pass the critical angle? But …
The other end of the rod is attached to a frictionless pivot.
… so one end is fixed, and so the mass has to move in a circle, doesn't it?

Doc Al
Mentor
But my question is, when the mass keep falling and the angle is larger than the critical angle, will the net force pointing outward so the acceleration is not pointing toward the center? If so, the object will not do a circular motion when it pass the critical angle?
Since, as tiny-tim points out, the mass is constrained to move in a circle, at all times the net force will point toward the center.

Passing the "critical point" just changes the rod's contribution to the net force from a push to a pull.