A pendulum is dropped 0.50 m from
the equilibrium position as shown below.
What is the speed of the pendulum bob as it passes through the
equilibrium position?

-- When the ball is at the top, before it's dropped, it has potential energy above the equilibrium position.
Potential energy = (mass) x (gravity) x (height) = (mass) x (G) x (0.5)
As it passes through the equilibrium position, it has kinetic energy.
Kinetic energy = (1/2) x (mass) x (speed)²
How much kinetic energy does it have at the bottom ? EXACTLY the potential energy that it started out with at the top !
THAT's where the kinetic energy came from. So the two expressions for energy are equal. K.E. at the bottom = P.E. at the top.
(1/2) x (mass) x (speed)² = (mass) x (G) x (0.5)
Divide each side by (mass) . . . (the mass of the ball goes away, and has no effect on the answer !)
(1/2) x (speed)² = (G) x (0.5)
Multiply each side by 2 :
(speed)² = G
speed = √G = √9.8 = 3.13 meters per second , regardless of the mass of the ball !

Answered by AL2006 | 2024-06-10

The speed of the pendulum bob as it passes through the equilibrium position can be calculated using conservation of energy principles. The mass cancels out, resulting in a final speed of approximately 3.13 m/s as it moves through the lowest point. This speed is derived from converting potential energy at the drop height into kinetic energy at the lowest point of the swing.
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Answered by AL2006 | 2024-10-01