- #1
Kernul
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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?