Question

"A problem involves a car of mass m going down a track from a height H, and round a loop of radius r. The loop is frictionless.

It asks for the minimum cut-off speed required, at the highest point in the loop (call it point D), such that the car makes it round the loop without falling. I know the solution; I should set the centripetal accleration equal to 9.81. In other words, contact force with the track at point D is equal to zero.

But I tried solving it by conservation of energy. At point D, the car is at a height 2r from ground level. Therefore, in order for the car to reach that height at point D, it must initially have a potential energy of mg(2r). Meaning, it should be released from a height H = 2r.

I got the wrong answer and I'm confused why that happened. Isn't that how conservation of energy work? Please clarify, where's the error in my solution?"

Answer 1

Step-by-step explanation:

At the topmost position, the car does not have zero velocity but it has velocity of v so that

v² /r = g or centripetal acceleration should be equal to g ( 9.8 )

Considering that, the car must fall from a height of 2r + h where

mgh = 1/2 mv²

= 1/2 m gr

So h = r/2

Hence the ball must fall from a height of

2r + r /2

= 2.5 r . So that it can provide velocity of v at the top where

v² / r = g .