Gravity and Rotation 4?

Question

What is the maximum speed with which a 1110 kg car can round a turn of radius 65 m on a flat road if the coefficient of static friction between tires and road is 0.80?

Answer in m/s.

Answer 1

** Edward is totally wrong. He does not know the correct formula for centrifugal force. My procedure is entirely correct.

In this case, if the centrifugal force is greater than the frictional force, the car will slide off. At maximum possible speed, the centrifugal force and the frictional force will be equal.

Here,
mass of the car, m = 1110 kg
radium, r = 65 m
coefficient of static friction, u = 0.8
gravitational acceleration, g = 9.8 m/s^2

centrigugal force = mv^2/r
frictional force = umg

now, mv^2/r = umg
or, v^2/r = ug
or, v^2 = ugr
or, v = (ugr)^1/2
or, v = (0.8 *9.8 * 65) ^1/2
so, v = 22.57 m/s

So the masimum speed of the car is 22.57 m/s (ans.)

Hope this was helpful.

Answer 2

Centripetal acceleration is due entirely to friction. From

a (cent.) = v² / r,

derive v² = ar = fr / m,

where f is friction force. In turn, f = µN = µmg. Thus,

v² = µmgr / m = µgr, and v = √µgr = 22.57 m/s, or about 81 kph.

Answer 3

Centripetal force must be equal or less than the force of static friction.

Fc < f
Fc = ma = .5 m v^2/R
f= u N= umg

now
.5 m v^2(1/R)= u m g

v=sqrt( 2 u g R)
v=sqrt(2 x 0.8 x 9.81 x 65.0)
v=32 m/s
actually
v< 32 m/s