Force exerted by rod on a mass moving in vertical circle

[jbriggs444]

Is this an action-reaction force? Which means there will be a force downward (with the same in magnitude as the force you have stated) that applies on the rod?

Yes. All real forces are "action-reaction" forces. If the rod pushes up on the mass, the mass pushes down on the rod. That is Newton's third law.

 

[Rikudo]

Not sure what you mean by "need" there.
We know it is radially inward because we are told the mass moves in a circle at uniform speed. Presumably this is arranged by a varying torque applied at the axle on which the rod rotates.

Is it not possible for the end of the rod to be the one which keep the speed constant? As what @vcsharp2003 is trying to say.

 

[haruspex]

Is it not possible for the end of the rod to be the one which keep the speed constant? As what @vcsharp2003 is trying to say.

I understand that question even less.
Some mechanism acts to keep the speed constant. We do not know and do not care what it is. We can deduce the force it must be exerting on the mass to achieve that, because we know that the net result of that force and mg is a radially inward force providing the centripetal acceleration.

 

[vcsharp2003]

Not sure what you mean by "need" there

I was just trying to decide whether $a_t>0$ or $a_t=0$, where $a_t$ stands for tangential acceleration in horizontal position. I wasn't trying to understand why speed is constant, which you've explained through a varying torque mechanism.

 

[haruspex]

I was just trying to decide whether $a_t>0$ or $a_t=0$, where $a_t$ stands for tangential acceleration in horizontal position. I wasn't trying to understand why speed is constant, which you've explained through a varying torque mechanism.

Tangential acceleration is the component of acceleration parallel to the velocity. If speed is constant, that must be zero.
(In vectors, $0=\frac{d}{dt}v^2=\frac{d}{dt}\vec v^2=2\vec v.\dot{\vec v}=2||\vec v|| a_t$.)

 

[vcsharp2003]

Tangential acceleration is the component of acceleration parallel to the velocity. If speed is constant, that must be zero.
(In vectors, $0=\frac{d}{dt}v^2=\frac{d}{dt}\vec v^2=\vec v.\dot{\vec v}=||\vec v|| a_t$.)

Yes, I get that part. If only direction of velocity is changing but not it's magnitude then tangential acceleration must be zero.

 

[vcsharp2003]

(In vectors, $0=\frac{d}{dt}v^2=\frac{d}{dt}\vec v^2=\vec v.\dot{\vec v}=||\vec v|| a_t$.)

Why did you take derivative of $v^2$?

 

[haruspex]

Why did you take derivative of $v^2$?

The speed, v, is constant if and only if $v^2$ is constant. Using the squared form let's me write it in vectors easily.
But I did omit a factor 2, now corrected.

 

[vcsharp2003]

The speed, v, is constant if and only if $v^2$ is constant. Using the squared form let's me write it in vectors easily.
But I did omit a factor 2, now corrected.

That's a very nice mathematical way of reasoning it, that I was unaware of. Thankyou for that great tip.