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?

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)`