Question

A marble slides without friction in a vertical loop around the inside of a smooth, 28.6 cm diameter horizontal pipe. The marble's speed at the bottom is 4.22 m/s and is fast enough so that the marble never loses contact with the pipe. What is its speed at the top?

Answer 1

Answer:3.49 m/s

Step-by-step explanation:

Given

Speed of marble at Bottom
`v=4.22 m/s`

Diameter of loop
`d=28.6 cm`

As Energy is conserved therefore Energy at top is equal to energy at bottom

`E_T=E_B`

`(mv^2)/(2)+mgh=(mv_0^2)/(2)` ,where
`v_0` is the velocity at bottom

`(v^2)/(2)+gh=(v_0^2)/(2)`

`v_0^2=v^2+2gh`

`v^2=v_0^2-2gh`

`v=√(v_0^2-2gh)`

`v=√(4.22^2-2* 9.8* 0.286)`

`v=√(17.8084-5.6056)`

`v=3.49 m/s`