Conic Pendulum Exercise: Tension and Velocity as Functions of Angle

[Kernul]

Homework Statement

A tether ball of mass $m$ is suspended by a rope of length $L$ from the top of a pole. A youngster
gives it a whack so that it moves with some speed $v$ in a circle of radius $r = L sin(\theta) < L$ around
the pole.
a) Find an expression for the tension $T$ in the rope as a function of $m$, $g$, and $\theta$.
b) Find an expression for the speed $v$ of the ball as a function of $\theta$.

Homework Equations

Centripetal acceleration
Newton's Second Law
Tension
Gravity

The Attempt at a Solution

This is the picture of how it would be with all the forces and with the y-axis upward and the x-axis going right. (I took this picture from another exercise about the conic pendulum in here)

Now, we know that there is no actual "Centripetal Force". The centripetal acceleration that the mass has is given by the net force of the other forces in play, in this case the tension $T$ and the gravity $m g$. We know that there is no motion on the y-axis and only a motion on the x-axis, so we have:
$$\begin{cases}
m a_x = T sin \theta \\
m a_y = T cos \theta - m g = 0
\end{cases}$$
We can write $a_x = a_c = \frac{v^2}{r}$.
We can then write:
$$T cos \theta = m g$$
$$T = \frac{m g}{cos \theta}$$
And so we have the tension expressed as requested.
The velocity, though, will be written like this:
$$v = \sqrt{a_c r}$$
$$v = \sqrt{a_c L sin \theta}$$
but we know that $a_c = \frac{T}{m} sin \theta$, $a_c = g tan \theta$, and so:
$$v = \sqrt{L g tan \theta sin \theta}$$
Am I right? Or I did something wrong?

 

[TSny]

Your work looks good to me. It's always a good idea to check limiting cases of your answer. For the cases θ = 0 and θ = 90^o, does your result for $v$ make sense? Also, does your formula for $v$ give the correct dimensions (or units) for a speed?

 

[Kernul]

Your work looks good to me. It's always a good idea to check limiting cases of your answer. For the cases θ = 0 and θ = 90^o, does your result for $v$ make sense? Also, does your formula for $v$ give the correct dimensions (or units) for a speed?

For the dimensions yes.
If I have $\theta = 0$, I have $v = 0$. This is the case where the ball is at the same height as the beginning of the rope. It nullifies because there's no force acting on it?
If I have $\theta = 90^o$, I have the tangent that doesn't exist. This is the case where the tether ball is basically at the center and so it doesn't actually move. Or, to better say, that the two forces acting on it nullify themselves.

 

[TSny]

For the dimensions yes.
If I have $\theta = 0$, I have $v = 0$. This is the case where the ball is at the same height as the beginning of the rope. It nullifies because there's no force acting on it?

θ = 0 corresponds to the rope being vertical.
θ = 90^o corresponds to the rope being parallel to the ground.

 

[Kernul]

θ = 0 corresponds to the rope being vertical.
θ = 90^o corresponds to the rope being parallel to the ground.

Oh true! I was looking at the other picture.
Thank you!