Problem 6-53 parts (a), (b), and (c) Centripetal force - Part 5 - A

A conical pendulum consists of a mass (the bob) on the end of a string. As the bob travels in a horizontal circle at constant speed, the string traces out a cone. For the pendulum shown, mass \(m = 1.95\; kg\), length \(l = 1.03\; m\), and angle \(\theta = 36.9^\circ\).

(a) Draw a free body diagram for the bob at the instant shown.
(b) What force constitutes the centripetal force?
(c) Determine the speed of the bob.

Accumulated Solution

\(T \cos \theta = mg\)

Correct!

\(T \sin \theta = mv^2/r(2) \)    This is the answer to part (b).

If you divide Eq (2) by Eq (1)

\(\frac{T \sin \theta}{T \cos \theta} = \frac{mv^2/r}{mg}\)

\(T\) and \(m\) both cancel leaving:

\(\frac{\sin \theta}{\cos \theta} = \frac{v^2}{rg}\)

In terms of \(l\) and \(\theta\), \(r\) is given by:

(A)   \(r = l \cos \theta\)

(B)   \(r = l \sin \theta\)