Question

To make a pendulum, a 210 g ball is attached to one end of a string that has a length of 1.2 m and negligible mass. (The other end of the string is fixed.) The ball is pulled to one side until the string makes an angle of 24° with the vertical; then (with the string taut) the ball is released from rest. Find (a) the speed of the ball when the string makes an angle of 12° with the vertical and (b) the maximum speed of the ball. (c) What is the angle between the string and the vertical when the speed of the ball is one-third its maximum value?

Answer 1

Answer:

a) $v \approx 1.233\,\frac{m}{s}$, b) $v \approx 1.426\,\frac{m}{s}$, c) $\theta_{2} \approx 22.61^{\textdegree}$

Explanation:

a) The speed of the ball is determined by applying the Principle of Energy Conservation:

$U_{g,1} + K_{1} = U_{g,2} + K_{2}$

The speed of the ball when the string makes an angle of 12° with the vertical is:

$K_{2} = U_{g,1} - U_{g,2} + K_{1}$

$\frac{1}{2}\cdot m \cdot v^{2} = m\cdot g \cdot L \cdot [(1-\cos \theta_{1})-(1 -\cos \theta_{2})]$

$v = \sqrt{2\cdot g \cdot L \cdot (\cos \theta_{2} - \cos \theta_{1})}$

$v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (1.2\,m)\cdot (\cos 12^{\textdegree}-\cos 24^{\textdegree})}$

$v \approx 1.233\,\frac{m}{s}$

b) The maximum speed of the ball is:

$v = \sqrt{2\cdot g \cdot L \cdot (\cos \theta_{2} - \cos \theta_{1})}$

$v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (1.2\,m)\cdot (\cos 0^{\textdegree}-\cos 24^{\textdegree})}$

$v \approx 1.426\,\frac{m}{s}$

c) The angle between the string and the vertical when the speed of the ball is one-third its maximum value is obtained by proving different values of $\theta_{2}$. The solution is approximately:

$\theta_{2} \approx 22.61^{\textdegree}$