Question

a circus man starts down an inclined plane on his scooter.At the end of the inclined plane,there is a vertical circular arch.If he is to safely negotiate the arch,what must be his velocity at the end of the plane?

Answer 1

The scooter must have a minimum velocity of $√(5gR)$ at the end of the plane.

Explanation:

Let

⇒ v₀ be the speed of the scooter at the end of the incline plane.

⇒ R be the radius of circular arc

⇒ v be the speed of scooter at highest point on the arc.

By Conservation of Energy

K.E. at lowest point = K.E. + P.E. at highest point

$\frac{1}{2}mv₀² =\frac{1}{2}mv² + mg(2R)$

$v₀² =v² + 4gR--------(1)$

By Balancing Force at Maximum Height of Arc

$mg + N = \frac{mv²}{R}$

$N = \frac{mv²}{R} - mg$

If the scooter must maintain contact with the arc

$N > 0$

$⇒ \frac{mv²}{R} - mg > 0$

$⇒ \frac{mv²}{R} >mg$

$⇒ v² >gR--------(2)$

Using this in (1), we get

$v₀² > gR + 4gR$

$v₀² > 5gR$

$v₀ > √(5gR)$

Therefore, the scooter must have a minimum velocity of $√(5gR)$ at the end of the plane.