**Applications of Centripetal Force**

**1. Motion of a car on a Level Curved Path**

Let F

_{1}and F_{2}be the forces of friction between the wheels and the road, directed towards the center of the horizontal curved track of radius r as shown in the figure. From the law of Friction, F

_{1}+ F_{2}= µ (R_{1}+ R_{2}) …... (i)Where µ is the coefficient of sliding friction between tires and the road and R

_{1}and R_{2}are the forces of normal reaction of the road on the wheels.As there is no motion of the car in the vertical direction, we have

R

_{1}+ R_{2}= m. g ….... (ii)As the force of friction provides the necessarily centripetal force, then

F

_{1}+ F_{2}= m. v^{2}/ r …...... (iii)So, From equation (ii) and (iii), we have

µmg = mv

^{2}/ r or, µ = v

^{2}/ rg .: v = √µ rg ….... (iv)

This equation gives the maximum velocity with which the car can take a circular turn of radius r and coefficient of friction µ. This shows that much more friction is required as the turning speed is to be increased and if the velocity is greater than determined by equation (iv), the skidding of the car occurs and moves out of the road. This equation holds for a vehicle of any mass from a bicycle to a heavy truck.

**2. Banking of road**

We can not always depend on the friction to move a car or other vehicles around a curve, especially if the road is icy or wet. In order not to depend on friction or to reduce wear on the tires on road, the roads are banked as shown in the figure.

The effect of banking is to tilt the normal toward the center of the curved road, so that it's inward horizontal component can supply the required centripetal force. So, resolving (R

_{1}+ R_{2}) into two components (R_{1}+ R_{2})sinθ towards the center and (R_{1}+ R_{2})cosθ where θ is the banking angle or angle of inclination of the plane with the horizontal, we have (R

_{1}+ R_{2}) sinθ = mv^{2}/rAnd vertically

(R

_{1}+ R_{2}) cosθ = mgFrom these two equations, we have

(R

_{1}+ R_{2}) sinθ/(R_{1}+ R_{2}) cosθ = mv^{2}/ r * 1/ mg or, Tanθ = v

^{2}/ rgThe above equation gives the angle for banking for a speed v and a curved track of horizontal radius r. so the tangent of the banking angle is directly proportional to the sped and inversely proportional to the radius of the path.

**3. Bending of a Cyclist**

When a cyclist takes a turn, he bends toward the center of the circular path and the horizontal component of normal reaction directed towards the center provides the necessary centripetal force for the motion. Suppose a cyclist of mass m going in a circular path of radius r with constant speed v tilting at an angle of θ with the vertical. Then R can be resolved into two rectangular components: R cosθ in the vertically upward direction and R sinθ, along the horizontal, towards the center of the circular track as shown in the figure. In equilibrium, R cosθ balances the weight mg of the cyclist and R sinθ provides the necessary centripetal force, mv

^{2}/ r.So,

R cosθ = mg …..... (i)

And,

R sinθ = mv

^{2}/ r …...... (ii)Dividing equation (ii) by equation (i), we get

Tanθ = v

^{2}/ rgThis relation gives the angle through which a cyclist tilts with the vertical while going in a circular road of radius r at a speed of v.

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