Circular Turning , Banking of Roads , Conical Pendulum

Circular Turning on Roads :

The necessary centripetal force while taking a circular turn is being provided to a vehicle by following three ways :

(i) by friction only

(ii) by banking of road only

(iii) by both friction and banking of roads

Friction only:

In this case the necessary centripetal force is provided by static friction given by

fs = µs mg

fs ≥ fc

µs mg ≥ mv2/r

v2 ≤ µs gr

$\large v \le \sqrt{\mu_s g r} $

For a given radius of curvature and coefficient of friction, the safe maximum velocity of the vehicle is given by

$\large v_{max} = \sqrt{\mu_s g r} $

Banking of Roads only :

Let θ be the angle through which the outer edge is raised relative to the inner edge. This angle is called “angle of banking”. The normal reaction N exerted by the road on the vehicle is directed normal to the surface as shown in the figure.

N cosθ balances the weight of the vehicle.

N cosθ = mg …(1)

N sinθ is directed towards the centre of the circular path. Which provides the centripetal force.

N sinθ = mv2/r ….(2)

From the above equations, we get

$\large tan\theta = \frac{v^2}{r g} $

$\large v = \sqrt{r g \; tan\theta}$

Motion of a vehicle on a Rough banked Road:

If friction is present between the road and the tyres, the components of friction and normal reaction provide the centripetal force.

Case I: If N sin θ > mv2/r, the vehicle possesses the tendency to slip down the plane. The minimum speed for avoiding slipping down the plane can be obtained by taking friction up the plane.

$\large N sin\theta – f cos\theta = \frac{m v_{min}^2}{r} $ …(i)

Ncosθ + fsinθ = mg ….(2) (f = μN )

From (1) and (2) we get

$\large v_{min} = \sqrt{\frac{r g(sin\theta – \mu cos\theta)}{cos\theta + \mu sin\theta}}$

$\large v_{min} = \sqrt{\frac{r g (tan\theta – \mu )}{1 + \mu tan\theta}}$

Case II: If N sinθ < mv2/r , the vehicle possesses the tendency to skid up the plane. The safe maximum speed for avoiding skidding can be obtained by taking friction acting down the plane.

$\large N sin\theta + f cos\theta = \frac{m v_{max}^2}{r} $ …(i)

Ncosθ – fsinθ = mg ….(2)

From (1) and (2) we get

$\large v_{max} = \sqrt{\frac{r g(sin\theta + \mu cos\theta)}{cos\theta – \mu sin\theta}}$

$\large v_{min} = \sqrt{\frac{r g (tan\theta + \mu )}{1 – \mu tan\theta}}$

Conical Pendulum:

A bob of mass M is given a horizontal push a little through angular displacement θ and arranged such that the bob describes a horizontal circle of radius ‘ r ’ with uniform angular velocity ω in such a way that the string always makes an angle θ with the vertical and T is the tension in the string.

Suppose the body is in rotational equilibrium, then

T cosθ = M g …(1)

T sinθ = M r ω2 …..(2)

From (1) and (2)

tanθ = r ω2/g

$\large \omega = \sqrt{\frac{g tan\theta}{r}}$

But r = l sinθ and ω = 2π/T0 (T0 is the time period of pendulum)

Time period of the pendulum is $\large T_0 = 2\pi \sqrt{\frac{l cos\theta}{g}}$

Solved Example: A small block is connected to one end of two identical massless strings of length 50/3 cm each with their other ends fixed to a vertical rod. If the ratio of tensions T1/T2 is 4 : 1, then what will be the angular velocity of the block? Take g = 9.8 ms-2.

Sol: For horizontal,

T1 sin 60° + T2 sin 60° = mω²r = mω²l sin 60°

⇒ T1 + T2 = mω²l …(i)

For vertical equilibrium of the block,

T1 cos 60° = T2 cos 60° + m g

⇒ T1 -T2 = mg/cos⁡60° = mg/(1/2) = 2 m g …(ii)

Dividing (i) by (ii),

$\large \frac{T_1 + T_2}{T_1 – T_2 }= \frac{\omega^2 l}{2 g} $

$\large \frac{T_1 /T_2 + 1}{T_1 /T_2 – 1 }= \frac{\omega^2 l}{2 g} $

⇒ $\frac{4+1}{4-1} = \frac{\omega^2 l}{2 g} $

$\large \omega^2 = \frac{10 g}{3 l} $

$\large \omega^2 = \frac{10 \times 9.8}{3 \times 50 \times 10^{-2}} $

or,  ω2 =196

or,  ω = 14 rad/s

Solved Example: A circular track has a radius of 10 m. If a vehicle goes round it at an average speed of 18 km/h, what should be the proper angle of banking.

Sol: We know angle of banking is given by

v = 18× (5/18) = 5 m/s

$\large tan\theta = \frac{v^2}{r g} = \frac{5^2}{10 \times 10}$

tanθ = 1/4

⇒ θ = tan-1 (1/4)

Q: Why are railway tracks and highways banked on the curves ?

Ans: When a road is banked , the normal reactions on the wheels give rise to component forces towards the center of the path , which is essential for the maintenance of the motion of a vehicle in a circular path . In the absence of banking the necessary centripetal force is provided by lateral pressure on the outer wheels . This involves wear and tear of tyres or wheels , in addition to chances of accidents at high speeds .

Q: A coin is put on the turntable of a record player . The motor is started , but before the final speed of rotation is reached , the coin flies off . Why?

Ans: In the beginning , the frictional force provides the necessary centripetal force to keep the coin on the circular path . With increasing speed , the frictional force also increases . This continues until limiting friction is reached , and exceeded . Then the coin flies off tangentially relative to an observer on the ground , the limiting frictional force now being unable to provide the necessary centripetal force .

Q: Can you examine whether a table top in a train is truly horizontal by using a spirit level ?

Ans: No , If the tracks are straight but inclined , the bubble shifts to one extreme position . The bubble shifts to one extreme position if the tracks are horizontal but curved . The bubble shifts to one extreme position if tracks are straight and level but the table top is not horizontal . Hence it can not be ascertained by a spirit level whether a table top in train is horizontal or not .

Q: A body moves along a circular path of radius 10 m and the coefficient of friction is 0.5. What should be its angular speed in rad/s if it is not to slip from the surface (g = 9.8 ms2)

(a) 5

(b) 10

(c) 0.1

(d) 0.7

Click to See Answer :
Ans: (d)


Q: The maximum speed that can be achieved without skidding by a car on a circular unbanked road of radius R and coefficient of static friction μ, is

(a) μRg

(b) $R g \sqrt{\mu }$

(c) $ \mu \sqrt{R g}$

(d) $\sqrt{\mu R g}$

Ans: (d)

Q: A car turns a corner on a slippery road at a constant speed of 10 m/s. If the coefficient of friction is 0.5, the minimum radius of the arc in meter in which the car turns is

(a) 20

(b) 10

(c) 5

(d) 4

Ans: (a)

Q: The maximum speed of a car a road turn of radius 30 m; if the coefficient of friction between the tyres and the road is 0.4; will be

(a)9.84 m/s

(b) 10.84 m/s

(c) 7.84 m/s

(d) 5.84 m/s

Ans: (b)

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