Question

A bead slides without friction around a loop the-loop. The bead is released from a height of 24.5 m from the bottom of the loop-the loop which has a radius 9 m. The acceleration of gravity is 9.8 m/s².

What is its speed at point A?
Answer in units of m/s.

Answer 1

`v = 11.29\ m/s`

Step-by-step explanation:

given,

bead is released from the height = 24.5 m from bottom

radius of the loop = 9 m

acceleration due to gravity = 9.8 m/s

A be the top most point of the loop

Difference of elevation from the top

H = 24.5 - 2 x r

H = 24.5 - 2 x 9

H = 24.5 - 18

H = 6.5 m

now using conservation of energy

KE = PE

`(1)/(2)mv^2 = m g H`

`v^2 = 2 g H`

`v = √(2 g H)`

`v = √(2 * 9.8 * 6.5)`

`v = √(127.4)`

`v = 11.29\ m/s`

speed at point A is equal to
`v = 11.29\ m/s`