A simple pendulum of length $1 m$ has a bob of mass $10 g$ . if the bob is released from the horizontal position, what will be its velocity at the lowest position? Take $g = 10 m s^{- 2}$ .

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Solution

Step 1: Given data,
Length of the pendulum, $L = 1 m$
Mass of the bob, $m = 10 g = 0.01 k g$
Step 2: Finding velocity at the lowest point,
The work done is equal to the loss of gravitational potential energy, $U_{g} = m g h$
Twice the length of the pendulum is equal to the height of the highest point from the lowest point, $h = L = 1 = 1 m$
Then the loss of potential energy of bob, $U_{g} = 0.01 \times 10 \times 1 = 0.1 J$
The loss of potential energy is equal to the gain in kinetic energy at the lowest point.
Then the kinetic energy at the lowest point, $K = U_{g} = 0.1 J$
The formula for kinetic energy, $K = \frac{1}{2} m v^{2}$
$\Rightarrow 0.1 = \frac{1}{2} \times 0.01 \times v^{2}$
$\Rightarrow v = \sqrt{20} \frac{m}{s}$
Hence, the velocity of the bob at the lowest position is $\sqrt{20} \frac{m}{s}$ .

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A simple pendulum of length 1m has a bob of mass 10g. if the bob is released from the horizontal position, what will be its velocity at the lowest position? Take g=10ms-2.

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