Force exerted by rod on a mass moving in vertical circle

In summary, (A) and (B) are obviously wrong but I think both (C) and (D) are correct. At the top, the forces acting on the mass are tension and weight, both directed downwards so the equation of motion will be: $$\text{Tension}+\text{Weight}=m.a$$ $$\text{Tension}=m.a-\text{Weight}$$ Based on that equation, tension will be minimum at the top position. The direction of tension is always towards the center as the mass moves in a circle. At the horizontal position, tension will always be towards the center to provide the centripetal force. At the bottommost position, it is also directed towards the center since the net force of
  • #36
haruspex said:
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.
 
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  • #37
haruspex said:
(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##?
 
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  • #38
vcsharp2003 said:
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.
 
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  • #39
haruspex said:
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.
 

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