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 in the previous problem is to be supplied by the normal force by the road what should be the proper angle of banking?

Answer

A, C

Explanation

IN the diagram
`Rcostheta=mg`…i
`Rsintheta=((Mv^2)/r)`……..ii
Dividing equation i with equation ii we get,
`tantheta =((mv^2)/(rmg))=v^2/(rg)`
Here `v=36 km/hr=10 m/sec, r=30m`
`:. Tanthete= v^2/(rg)=100/(30xx10)=(1/3)`
`rarr thetat=tan^-1 (1/3)`