The string in the figure is 50 cm long. When the ball is released from rest, it swings along the dotted arc. How fast is it going at the lowest point in its swing?
A) 2.0 m/s
B) 2.2 m/s
C) 3.1 m/s
D) 4.4 m/s
E) 6.0 m/s

Question

The string in the figure is 50 cm long. When the ball is released from rest, it swings along the dotted arc. How fast is it going at the lowest point in its swing? 
A) 2.0 m/s 
B) 2.2 m/s 
C) 3.1 m/s 
D) 4.4 m/s 
E) 6.0 m/s

Quizplus · Accepted Answer

Using conservation of energy, we can find the speed of the ball at the bottom of the swing. At the top of the swing, all of the energy is in the form of potential energy, so:

$mg\frac{L}{2}=mgh$

where $h$ is the height of the swing, $L$ is the length of the string, and $g$ is the acceleration due to gravity. Solving for $h$ gives:

$h=\frac{L}{2}-\frac{L}{2}\cos	heta=L\left(\frac{1}{2}-\frac{1}{2}\cos	hetaight)$

where $	heta$ is the angle the string makes with the vertical at the bottom of the swing. At the bottom of the swing, all of the energy is in the form of kinetic energy, so:

$\frac{1}{2}mv^2=mgh$

Substituting for $h$, we get:

$\frac{1}{2}mv^2=mgL\left(\frac{1}{2}-\frac{1}{2}\cos	hetaight)$

Solving for $v$, we get:

$v=\sqrt{gL(1-\cos	heta)}$

At the bottom of the swing, $\cos	heta=-1$, so:

$v=\sqrt{2gL}$

Plugging in numbers, we get:

$v=\sqrt{2	imes 9.81	imes 0.5}=3.13 \approx \boxed{	extbf{(C) }3.1 	ext{ m/s}}$

The String in the Figure Is 50 Cm Long