A car of mass 'm' is moving on a horizontal circular path of radius r. At an instant its speed is 'v' and is increasing at a rate 'a'.
(a) the acceleration of the car is towards the center of the path.
(b) the magnitude of the frictional force on the car is greater than mv²/r.
(c) the friction coefficient between the ground and the car is not less than a/g.
(d) the friction coefficient between the ground and the car is µ=tan-1(v²/rg).
Answers
Answer ⇒ Option (b). and Option (d).
Explanation and Mathematical Proof ⇒ Since, the car is moving in circular path and its velocity's magnitude is changing and obviously the direction of velocity will change, this means that there will be two component of the acceleration, one parallel to the path and other Perpendicular to the path. I means there will be Centripetal acceleration and tangential acceleration.
Now, the Net acceleration produced by them will not be along the Centre. Hence, Option (a). is incorrect.
The magnitude of the Frictional force must be less than the mv²/r so that car can remains in the path. Therefore, Option (b). is correct.
Also, f ≥ mv²/r
⇒ μmg ≥ mv²/r
⇒ μg ≥ a
⇒ μ ≥ a/g.
This tells us that coefficient of the Friction cannot be less than a/g. Hence, Option (c). is correct.
But option (d). can never be correct. It is just the case of angle of banking and cannot be valid here.
Hope it helps.