Page 118 - Class-11-Physics-Part-1_Neat

104 PHYSICS

5.10 CIRCULAR MOTION is the static friction that provides the centripetal
acceleration. Static friction opposes the
We have seen in Chapter 4 that acceleration of
impending motion of the car moving away from
a body moving in a circle of radius R with uniform the circle. Using equation (5.14) & (5.16) we get
2
speed v is v /R directed towards the centre.
the result
According to the second law, the force f providing
c 2
this acceleration is : mv
f = ≤ µ s N
mv 2 R
f = (5.16)
c
R µ RN
2
v ≤ s = µ s Rg [∵N = mg]
where m is the mass of the body. This force m
directed forwards the centre is called the which is independent of the mass of the car.
centripetal force. For a stone rotated in a circle This shows that for a given value of µ and R,
s
by a string, the centripetal force is provided by there is a maximum speed of circular motion of
the tension in the string. The centripetal force the car possible, namely
for motion of a planet around the sun is the
v = µ Rg (5.18)
max s

(a) (b)

Fig. 5.14 Circular motion of a car on (a) a level road, (b) a banked road.

gravitational force on the planet due to the sun. Motion of a car on a banked road
For a car taking a circular turn on a horizontal
We can reduce the contribution of friction to the
road, the centripetal force is the force of friction.
circular motion of the car if the road is banked
The circular motion of a car on a flat and
(Fig. 5.14(b)). Since there is no acceleration along
banked road give interesting application of the
the vertical direction, the net force along this
laws of motion.
direction must be zero. Hence,
Motion of a car on a level road
N cos θ = mg + f sin θ (5.19a)
Three forces act on the car (Fig. 5.14(a):
(i) The weight of the car, mg The centripetal force is provided by the horizontal
(ii) Normal reaction, N components of N and f.
(iii) Frictional force, f 2
mv
As there is no acceleration in the vertical N sin θ + f cos θ = (5.19b)
R
direction
N – mg = 0 But f µ≤ s N
N = mg (5.17)
Thus to obtain v we put
The centripetal force required for circular motion max
is along the surface of the road, and is provided f = µ s N .
by the component of the contact force between
Then Eqs. (5.19a) and (5.19b) become
road and the car tyres along the surface. This
by definition is the frictional force. Note that it N cos θ = mg + µ s N sin θ (5.20a)

2018-19