Question

You're driving a vehicle of mass 850 kg and you need to make a turn on a flat road. The radius of curvature of the turn is 80 m. The maximum horizontal component of the force that the road can exert on the tires is only 0.22 times the vertical component of the force of the road on the tires (in this case the vertical component of the force of the road on the tires is , the weight of the car, where as usual 9.8 N/kg, the magnitude of the gravitational field near the surface of the Earth). The factor 0.22 is called the "coefficient of friction" (usually written "", Greek "mu") and is large for surfaces with high friction, small for surfaces with low friction.(a) What is the fastest speed you can drive and still make it around the turn? Invent symbols for the various quantities and solve algebraically before plugging in numbers.

maximum speed = ____ m/s
(b) Which of the following statements are true about this situation?
The net force is nonzero and points away from the center of the kissing circle.The momentum points toward the center of the kissing circle.The net force is nonzero and points toward the center of the kissing circle.The rate of change of the momentum is nonzero and points toward the center of the kissing circle.The centrifugal force balances the force of the road, so the net force is zero.The rate of change of the momentum is nonzero and points away from the center of the kissing circle.
(c) Look at your algebraic analysis and answer the following question. Suppose your vehicle had a mass 3 times as big (5250 kg). Now what is the fastest speed you can drive and still make it around the turn?
maximum speed = ____ m/s
(d) Look at your algebraic analysis and answer the following question. Suppose you have the original 1750 kg vehicle but the turn has a radius twice as large (166 m). What is the fastest speed you can drive and still make it around the turn?
maximum speed = ____m/s

Answer 1

(a) v = 13.13 m/s

(b) The centrifugal force balances the force of the road, so net force is zero.

(c) v = 13.13 m/s

(d) v = 18.92 m/s

Step-by-step explanation:

(a)

To make it around the turn without skidding the frictional force on cat must balance the centrifugal force. Therefore:

Frictional Force = Centrifugal Force

μR = mv²/r

where,

R = Normal Reaction = Weight of Car = mg

Therefore,

μmg = mv²/r

μg = v²/r

v = √μgr

where,

v = maximum possible velocity of car = ?

μ = coefficient of friction = 0.22

g = 9.8 m/s²

r = radius of curvature = 80 m

Therefore,

v = √[(0.22)(9.8 m/s²)(80 m)

v = 13.13 m/s

(b)

In order for the car to move without skidding around the turn, all the forces in horizontal direction must be equal. Hence, the centrifugal force and the frictional force (force of the road) must balance each other. So the true statement is:

The centrifugal force balances the force of the road, so net force is zero.

(c)

v = √μgr

Since the formula for speed is independent of mass. Therefore, the speed will remain same.

v = 13.13 m/s

(d)

v = √μgr

v = √[(0.22)(9.8 m/s²)(166 m)

v = 18.92 m/s