Question

What is the greatest speed at which a car having a track of 1.5m can turn in a circular path of radius 24.5m without overturning assume that the centre of gravity of the car is 0.49m above the ground

Answer 1

Given:

Track of car= 1.5m

Radius of circular path= 24.5m

Centre of gravity is 0.49m above the ground

To find:

Max speed at which car can travel without overturning.

Solution:

• Two forces act on the body. Its weight (mg) acts downward and centrifugal force acts outward.
• To avoid overturning, torque due to centrifugal force should always be  lesser than or equal to torque due to weight.
• At maximum velocity, torque due to centrifugal force= torque due to mg.
• ∴ mg * (distance from axis) =( mv²/r) * (distance from axis).
• g * 0.75m = v² *0.49/R
• ∴ $v_{max} = \sqrt{ \frac{9.8*0.75*24.5}{0.49} }$  
• ∴ v = 19.2 m/sec

Answer:

Maximum velocity without overturning will be 19.2 m/sec.