Applications of Centripetal Force | Physics Grade XI

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Applications of Centripetal Force | Physics Grade XI

Applications of Centripetal Force

Applications of Centripetal Force

1. Motion of a car on a Level Curved Path
Let F1 and F2 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,
     F1 + F2 = µ (R1 + R2) …... (i)
Where µ is the coefficient of sliding friction between tires and the road and R1 and R2 are the forces of normal reaction of the road on the wheels.
Motion of car on level curved pathAs there is no motion of the car in the vertical direction, we have
     R1 + R2 = m. g ….... (ii)
 
As the force of friction provides the necessarily centripetal force, then
     F1 + F2 = m. v2/ r …...... (iii)
 
So, From equation (ii) and (iii), we have
     µmg = mv2/ r
     or, µ = v2/ 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. 
Banking of roadThe 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 (R1 + R2) into two components (R1 + R2)sinθ towards the center and (R1 + R2)cosθ where θ is the banking angle or angle of inclination of the plane with the horizontal, we have
     (R1 + R2) sinθ = mv2/r
 
And vertically
     (R1 + R2) cosθ = mg
 
From these two equations, we have
     (R1 + R2) sinθ/(R1 + R2) cosθ = mv2/ r * 1/ mg
     or, Tanθ = v2/ rg
 
The 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, mv2/ r.
Bending of cyclistSo,
     R cosθ = mg …..... (i)
 
And,
     R sinθ = mv2/ r …...... (ii)
 
Dividing equation (ii) by equation (i), we get
     Tanθ = v2/ rg
This 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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