Question

For the rest of the problems involving this pendulum, m = 20 kg and l = 20 m. Suppose the pendulum is at the top, moving at 12 m/s horizontally. How fast will it go at the bottom? 13 m/s 23 m/s 31 m/s 52 m/s We can't tell. The ball isn't moving fast enough at the top for a cable tension to exist, and so just falls freely (like a freely-falling body) until it's yanked by the cable somewhere near the bottom. Suppose instead, the pendulum is moving 16 m/s horizontally at the top. How fast will it go at the bottom?

Answer 1

Here ,

m = 20 Kg

l = 20 m

initial speed at the top , u = 12 m/s

at the bottom of the path

let the speed of pendulum is v

Using conseravtion of energy

0.5 * m * v^2 = 0.5 * m * u^2 + m *g * (2 * l)

0.5 * v^2 = 0.5 * 12^2 + 2 * 9.8 * 20

solving for v

v = 30.5 m/s = 31 m/s

the speed of the ball at the bottom is 31 m/s

20)

for u = 16 m/s

Using conseravtion of energy

0.5 * m * v^2 = 0.5 * m * u^2 + m *g * (2 * l)

0.5 * v^2 = 0.5 * 16^2 + 2 * 9.8 * 20

solving for v

v = 32 m/s = 32 m/s

the speed of the ball at the bottom is 32 m/s