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

Answer:

 $v = 11.29\ m/s$

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

 $\dfrac{1}{2}mv^2 = m g H$

 $v^2 = 2 g H$

 $v = \sqrt{2 g H}$

 $v = \sqrt{2 \times 9.8 \times 6.5}$

 $v = \sqrt{127.4}$

 $v = 11.29\ m/s$

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