Centripetal Force
Q1

The orbit of the moon about the earth, which is nearly circular has a radius of about 384000 km and a period T of 27.3 days. Determine the acceleration of the moon towards the earth.?

Sol:
Radius of the orbit r = 384000 km
; = 3.84 × 108 m
Time required for one complete orbit is
T = 27.3 days
= 27.3 × 24 × 3600
= 2.36 × 106 seconds.
velocity of the moon moving around the earth
v =
=
=
= 10.22 × 102 m/s
The centripetal acceleration of the moon
a =
=
= 27.2 × 10- 4 m/s2
= 2.72 × 10- 3 m/s2
Q2

At what maximum speed can a car safely negotiate a horizontal unbanked turn of radius 60m in dry weather if the coefficient of static friction is 0.95?

Sol:

The speed v at which the car of mass m can negotiate a turn of radius r is related to the centripetal force Fc

In this case the maximum static frictional force Fs provides the centripetal force.
we know that Fs = μs FN,
FN is the normal force and μs is the coefficient of static friction.

As there is no motion in vertical direction, the weight balances the normal force
FN = mg

Q3

A cyclist goes around a circular track 500m long in 25 seconds. What is the angle that his cycle makes with the vertical to safely cover the circular track ?

Sol:

Given that the track is 500m long i.e., perimeter of the circle
2πr = 500m
r =
r = 80m

Time taken by the cyclist to complete one round
T = 25sec

In order to take a safe turn, the cyclist has to bend a little from his vertical position

When the cyclist bends, the parallel component of the normal force provides the required centripetal force.
The perpendicular component is equal to the weight of the cyclist, as there is no motion in perpendicular direction.
FN cos θ = mg
FN sin θ =

Now,

Now we need to find the velocity to determine the safe angle of the cyclist

Velocity v =
=
; = 20 m/s

θ = 27° is the angle at which the cyclist should bend to safely cover the track.

Q4

Two curves on a highway have the same radii. However, one is banked at angle θ and the other is unbanked. A car can safely travel along the unbanked curve at a maximum speed v0 under conditions when the coefficient of static friction between the tyres and the road is μs = 0.81. The banked curve is frictionless, and the car can negotiate it at the same maximum speed v0. Find the angle θ of the banked curve?

Sol:

Let r be the radii of the two curves. v0 is the maximum speed.
When the car travels along the unbanked curve.
The maximum static frictional force provides the necessary centripetal force

Fs = Fc
FN = Normal force and is equal to the weight of the car = mg

Now, when the car travels along the banked curve, the angle of the curve is given by

θ = 39° is the angle of banking