JEE Advanced2020PhysicsLaws of MotionActual
A student skates up a ramp that makes an angle 30 ° with the horizontal. He/she starts (as shown in the figure) at the bottom of the ramp with speed v 0 and wants to turn around over a semicircular path x y z of radius R during which he/she reaches a maximum height h (at point y ) from the ground as shown in the figure. Assume that the energy loss is negligible and the force required for this turn at the highest poin
Options
- Av 0 2 - 2 g h = 1 2 g R
- Bv 0 2 - 2 g h = 3 2 g R
- Cthe centripetal force required at points x and z is zero
- Dthe centripetal force required is maximum at points x and z
Correct answer
A. v 0 2 - 2 g h = 1 2 g R
Step-by-step solution
Speed at x and z is equal and also maximum in circular track. From energy conversation, 1 2 m v 0 2 = m g h + 1 2 m v 2 .......(1) at y ,   mg sin 30 ° = m v 2 R ⇒ v 2 = R g 2 ∴ From equation (1) v 0 2 - 2 g h = R g 2