Question

A scooter weighing 150 kg together with its rider moving at 36 km/hr is to take a turn of radius 30 m. What horizontal force on the scooter is needed to make the turn possible?
If the horizontal force needed for the turn is to be supplied by the normal force by the road, what should be the proper angle of banking?

Answer 1

Mass, m=150 kg ,  

Speed, v=36  

km/hr =36000 m/3600  

s =10 m/s  

Radius, r = 30 m    

Force needed to make the turn possible = mv²/r    =150x10x10/30 = 500 N..

Answer 2

Given conditions ⇒

Velocity =  36 km/hr = 10 m/s.  Mass = 150 kg.

Radius = 30 m.

Horizontal Force needed must be equal to the centripetal force so that Scotter can take turn without sliding.

∴ Horizontal Force = mv²/r

= 150 × 100/30

= 500 N.

Now,

If the Horizontal Force should be supplied by the Normal Force then,  

NSinθ = mv²/r

NCosθ = mg

Dividing both equations,

tan θ = v²/rg

⇒ tan θ = 100/10 × 30

⇒ tan θ = 1/3

⇒  θ = tan⁻¹(1/3).

Hope it helps.