The question has asked us to find the velocity and tension in the string whose lenght is L. Let 'r' be the radius of the circular path of the bob. I think the free body diagram looks like this

enter image description here

I want to calculate tension first

In the diagram , T = mg.costheta and mg = T.costheta

From the driagram of the entire system

costheta = √(L^2 - r^2) / L

So my answer depends on which equation i use. The second one is the correct one. How to know which equation we have to use? the answer for velocity is is also different for the two equations.

Is there something like i should only resolve one of these two? Because since tension's component is what i need to find centripetal force, i dont really need to resolve mg?


closed as off-topic by John Rennie, glS, stafusa, Yashas, Jon Custer Dec 12 '17 at 20:48

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – John Rennie, glS, stafusa, Yashas, Jon Custer
If this question can be reworded to fit the rules in the help center, please edit the question.


Hope this helps

enter image description here

So you have:

ΣFy = T cosθ = mg (since the mass is stationary in the y-direction)

ΣFx = T sinθ = ma = $\frac{mv^2}{r}$ (since the sine component is responsible for the centripetal force of the circular motion)

From pythagoras you know that sinθ = $\frac{r}{L}$ so

T sinθ = T$\frac{r}{L}$ = $\frac{mv^2}{r}$

By dividing T$\frac{r}{L}$ = $\frac{mv^2}{r}$ by T cosθ = mg you get

$\frac{r}{cos θ L}$ = $\frac{v^2}{g r}$

By rearranging the above equation:

v = r ⋅ sqrt($\frac{g}{cosθ L}$), which is the tangetial speed of the mass.

The angular speed ω = $\frac{v}{r}$ so

ω = sqrt($\frac{g}{cosθ L}$)

The tension in the string can either be written as

T = $\frac{m v^2 L}{r^2}$


T = $\frac{m g}{cosθ}$

Hope it helped

  • $\begingroup$ Hi and welcome to the Physics SE! Thanks for your contribution. Please note, tough, that according to the current Homework policy, (near-)complete answers are discouraged and might be deleted. $\endgroup$ – stafusa Dec 12 '17 at 9:28
  • $\begingroup$ Thanks for the reply. Your answer is right . After getting the equation Tsin(theta) =mv^2/r and substituting the value of sin(theta) , youve divided the equation by Tcos(theta) = mg . What i did was that i didnt use this second equation instead resolved mg to get mg.cos(theta) = T and then substituted for T in the Tsin(theta) = mv^2/r which gave me mg.cos(theta).sin(theta) = mv^/r and then substituted the values for cos(theta) and sin(theta) . Can you also please tell why this is a wrong method? What i think is that we should resolve the quantity that we are asked to find . Is that the case? $\endgroup$ – user178392 Dec 12 '17 at 9:34
  • $\begingroup$ @stafusa thank you very much. I agree the question is pretty much just beneficial and interesting to me so please feel free to delete it because i cant :(. I'll keep the guidelines in mind in the future. $\endgroup$ – user178392 Dec 12 '17 at 12:09
  • $\begingroup$ @user178392, no big deal. I'm no moderator, so I can't singlehanded delete posts, and I don't think I would either, since I'm usually not flagging answers for this reason (even if I do vote to close some homework questions). $\endgroup$ – stafusa Dec 12 '17 at 14:39
  • $\begingroup$ @user178392 The error you made was the assumption that mg has components in both the x and y direction while in reality (from my drawing) only has a component in the -y direction. By assuming that the mg has an x and a y component (as is the case in a pendulum in simple harmonic motion) you imply that the pendulum is swinging while simultaneously rotating in circular motion and this is not the case (if that makes sense). $\endgroup$ – A.Wichink Kruit Dec 12 '17 at 14:56

Not the answer you're looking for? Browse other questions tagged or ask your own question.