Question 197: Work, Energy and Power & Circular Motion

Question

A pendulum bob has a speed $4 \mathrm{~m} / \mathrm{s}$ at its lowest position. The pendulum is 1 m long. When the length of the string makes an angle of $60^{\circ}$ with the vertical, the speed of the bob at that position is (acceleration due to gravity, $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$, $\cos 60^{\circ}=0.5$ )

Accepted Answer

Step 1. Known values:
Length of pendulum: $L = 1 \, 	ext{m}$
Speed at lowest position: $v_{	ext{lowest}} = 4 \, 	ext{m/s}$
g = $10 \, 	ext{m/s}^2$
Position of interest: string at $60^\circ$ with vertical.
Step 2. Use energy conservation
At lowest point, potential energy is minimum.
At another position, some kinetic energy is converted into potential energy:
$ \frac{1}{2} m v^2 + m g h = \frac{1}{2} m v_{	ext{lowest}}^2 $
Here $h$ is the rise in height of bob above lowest position.
Step 3. Calculate vertical rise $h$
At lowest: vertical position = $-L) $.
At angle $	heta = 60^\circ$, vertical height relative to pivot = $-L \cos	heta = -L \cos 60^\circ$.
So rise:
$ h = L - L \cos 	heta = L(1 - \cos	heta) $
$ h = 1 	imes (1 - 0.5) = 0.5 \, 	ext{m} $
Step 4. Apply energy conservation
$ \frac{1}{2} m v^2 + m g (0.5) = \frac{1}{2} m (4^2) $
$ \frac{1}{2} v^2 + 5 = \frac{1}{2}(16) = 8 $
$ \frac{1}{2} v^2 = 3 $
$ v^2 = 6 \quad \Rightarrow \quad v = \sqrt{6} \, 	ext{m/s} $
Ã¢ÂÂ Final Answer:
The speed of the bob at the $60^\circ$ position is:
Option C: $\sqrt{6} \, 	ext{m/s}$

Answer

$6 \mathrm{~m} / \mathrm{s}$

Answer

$\sqrt{3} \mathrm{~m} / \mathrm{s}$

Answer

$\sqrt{6} \mathrm{~m} / \mathrm{s}$

Answer

$3 \mathrm{~m} / \mathrm{s}$

Work, Energy and Power & Circular Motion

Choose the correct answer: