Question

A simple pendulum consisting of a small heavy bob attached to a light string of length 40cm is released from rest with the string at 60° to the downward vertical. Find the speed of the pendulum bob as it passes through its lowest point.

Answer 1

Answer on Question #49706, Physics, Mechanics | Kinematics | Dynamics

A simple pendulum consisting of a small heavy bob attached to a light string of length $40\mathrm{cm}$ is released from rest with the string at $60{}^{\circ}$ to the downward vertical. Find the speed of the pendulum bob as it passes through its lowest point.

Solution:

The total mechanical energy (ME) of a body, is the sum of its kinetic energy (KE) and its gravitational potential energy (PE):

$$M E = K E + P E = c o n s t a n t$$

Thus,

$$K E _ {f i n a l} = P E _ {i n i t i a l}$$ $$\frac {m v ^ {2}}{2} = m g h$$ $$v = \sqrt {2 g h}$$ $$h = L - L \cos (\theta_ {0})$$

Hence,

$$v = \sqrt {2 g L (1 - \cos (\theta_ {0}))} = \sqrt {2 * 9 . 8 * 0 . 4 (1 - \cos (6 0 {}^ {\circ}))} = 1. 9 8 \mathrm {m / s}$$

Answer: $v = 1.98 \, \text{m/s}$

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