# Circular motion, deciding what the tangential speed must be to maintain theta [closed]

A bob of mass m = 0.250kg is suspended from a fixed point with a massless string of length L = 25.0cm . You will investigate the motion in which the string traces a conical surface with half-angle θ = 21.0 deg So what is the relationship between theta, mass and Tangential speed? An equation and an explanation would be nice as my text book is somewhat lacking when it comes to circular motion it only deals with orbit and im not sure if I can change one of those equations for this problem.

• For more info : conical pendulum Mar 2, 2014 at 6:24
• Yikes, I was googling the incorrect term I guess thank ou both for your answers. Sandeep thank you your equations worked well and helped answer my next few questions too. Mar 2, 2014 at 6:26

The string is essentially determining the radius of the uniform circular motion. So you just use the length of the string and the angle to find the radius of rotation (i.e. distance from particle to black dot). The relationship is just from right triangle trig

$$\sin \theta = r/L$$

Then using Newtons laws with $a=v^2/r$ we get

$$F=mv^2/r=m\frac{v^2}{L\sin\theta}$$

I just noticed that @Sandeep Thilakan answered nicely (upvote from me!) the question before me :( but I will just make it explicit where he got Eq. 1 and 2 from.

Equation 1: From the x components of the force vectors (Tension of string). There is only acceleration in the x and is centripetal so it is always pointed towards the center.

\begin{align}\sum F_{x}=&ma_{x}\\\ T\sin\theta =& mv^2/r \\\ T\sin^2\theta =& mv^2/L\end{align}

Equation 2: No acceleration in the y direction so we get

\begin{align}\sum F_{y} =& ma_{y} \\\ T\cos\theta -mg =& 0 \\\ T\cos\theta =&mg \longrightarrow T=mg/\cos\theta\end{align}

Putting these together we get

$$mg\sin^2\theta / \cos\theta = mv^2/L \longrightarrow \sin^2\theta/\cos\theta = \frac{v^2}{gL}$$

And as Sandeep pointed out if we use $v=\omega r=\omega L\sin\theta$ we get a nicer answer of

$$1/\cos\theta=\frac{\omega^2 L}{g}$$

• Thank you for expounding on the answer. Definitely gives a more conceptual understanding. Mar 2, 2014 at 6:46

From the FBD, since there is no vertical acceleration we get,

$mg = T \cos \theta \tag{1}$ where $T$ is the tension in the string

From the circular motion of the bob, we get

$\frac{mv^2}{L \sin \theta} = T \sin \theta \tag{2}$

Simplifying $(1)$ and $(2)$

$\tan \theta \sin \theta = \frac{v^2}{Lg}$

Infact, if you use angular velocity $\omega$ instead of tangential velocity $v$, you get a simpler expression

$\cos \theta = \frac{g}{L \omega ^2}$