An expression for the maximum safe speed for a vehicle on a circular horizontal road without toppling/overturning/rollover :

Consider a car of mass m taking a turn of radius r along a level road. As seen from an inertial frame of reference, the forces acting on the car are :

The lateral limiting force of static friction $\vec{f_s}$ on the wheels acting along the axis of the wheels and towards the centre of the circular path which provides the necessary centripetal force see Fig.

The weight $vec{m_g}$ acting vertically downwards at the centre of gravity (C.G.)

The normal reaction $\vec{N}$  of the road on the wheels, acting vertically upwards effectively at the C.G.

Since maximum centripetal force = limiting force of static friction,

ma_r  = $\frac{mv^2}{r} = f_s$          …..(1)

In a simplified rigid-body vehicle model, we consider only two parameters-the height h of the C.G. above the ground and the average distance ‘w’ between the belt and right wheels called the track width.

The friction force $\vec{f_s}$ on the wheels produces a torque τ_t that tends to overturn/rollover the car about the outer wheel. Rotation about the front-to-back axis is called roll.

τ_t = f_s.h = $(\frac{mv^2}{r})h$ …….. (2)

When the inner wheel just gets lifted above the ground, the normal reaction $\vec{N}$ of the road acts on the outer wheels but the weight continues to act at the C.G. Then, the couple formed by the normal reaction and the weight produces a opposite torque τ_r , which tends to restore the car back on all four wheels ref. Fig.

τ_r =mg. $(\frac{w}{2})$  …….(3)

The car does not topple as long as the restoring torque τ_t  counterbalances the toppling torque τ_t

Thus, to avoid the risk of rollover, the maximum speed that the car can have is given by

$(\frac{mv^2}{r})h$  = mg.$(\frac{w}{2})$ 

∴$v_{max}=\sqrt{\frac{rwg}{2h}$      ……(2)

Thus, vehicle tends to roll when the radial acceleration reaches a point where inner wheels of the four-wheeler are lifted off of the ground and the vehicle is rotated outward. A rollover occurs when the gravitational force $\vec{mg}$ passes through the pivot point of the outer wheels, i.e., the C.G. is above the line of contact of the outer wheels. Equation (3) shows that this maximum speed is high for a car with larger track width and lower centre of gravity.

There will be rollover (before skidding) if τ_t ≥ τ_r  that is if

f_s.h ≥ mg.$(\frac{w}{2})$   

i.e., if 

$\frac{f_s}{mg} ≥ \frac{w}{2}$ or $\frac{a_r}{g} ≥ \frac{w}{2h}$ or $\frac{f_s}{μ_s} ≥ \frac{w}{2h}$

The vehicle parameter ratio $\frac{w}{2h}$ is called the static stability factor (SSF). Thus, the risk of a rollover is low if SSF ≤ μ_s.  A vehicle will most likely skid out rather than roll if m_s is too low, as on  wet or icy road.

Expression for the optimum or most safe speed with which a vehicle can be driven along a curved banked road.

Fig. Optimum (most safe) speed on a banked road

Consider a car taking a left turn along a road of radius r banked at an angle θ for a designed optimum or most safe speed v_o. Let m be the mass of the car. In general, the forces acting on the car are

(i) its weight mg acting vertically down

(ii) the normal reaction of the road N perpendicular to the road surface

(iii) the frictional force f_s along the inclined surface of the road.

At the optimum speed, frictional force is not relied upon to contribute to the necessary lateral centripetal force.

Thus, ignoring f_s , resolve N into two perpendicular components :

N cosθ vertically up and N sinθ horizontally towards the centre of the circular path.

Since there is no acceleration in the vertical direction, N cosθ balances mg and N sinθ  provides the necessary centripetal force.

N sinθ = $\frac{mv_0^2}{r}$    …..(1)

N cosθ = mg       …..(2)

∴ $\frac{N sinθ}{N cosθ}=\frac{\frac{mv_0^2}{r}}{mg}$

∴ tanθ = \{\frac{v_o^2}{rg}\)  or θ =tan^-1 $\frac{v_0^2}{rg}$ …..(3)

Equation (3) gives the expression for the required angle of banking.

From Eq. (3), we can see that

• angle θ depends upon v_o, r and g.

• The angle of banking is independent of the mass of a vehicle negotiating the curve.

Also, for a given r and θ, the recommended optimum speed is

v_o = $\sqrt{rgtanθ}$

Expression for (a) the minimum safe speed (b) the maximum safe speed with which a vehicle can negotiate the curve without skidding on a banked circular road which is designed for traffic moving at an optimum or most safe speed v_o :

Suppose a car taking a left turn along a road of radius r banked at an angle θ for a designed optimum or most safe speed v_o. Let m be the mass of the car. In general, the forces acting on the car are

(i) its weight mg acting vertically down

(ii) the normal reaction of the road N perpendicular to the road surface

(iii) the frictional force f_s along the inclined surface of the road.

If μ_s is the coefficient of static friction between the tyres and road, f_s = μ_sN

(a) For minimum safe speed : If the car is driven at a speed less than the optimum speed v_o, it may tend to slide down the inclined surface of the road so that

f_s is up the incline, see Fig.

Fig. Minimum safe speed on a banked road

Resolve N and f_s into two perpendicular components : N cosθ and f_s sinθ vertically up; N sinθ horizontally towards the centre of the circular path while f_s cosθ

horizontally outward. So long as the car takes the turn without sliding down, the sum

N cosθ + f_s sinθ balances mg, and N sinθ — f_s cosθ provides the necessary centripetal force. If v_min is the minimum safe speed without skidding

$\frac{v^2_{min}}{r}$ = N sinθ — f_s cosθ = N(sinθ — μ_scosθ) ………..(1)

And mg = N cosθ + f_s sinθ = N(cosθ + μ_ssinθ)  ………..(2)

∴ Dividing Eq. (1) by Eq- (2)

∴ $\frac{v^2_{min}}{rg}$ = $\frac{sinθ-μ_s cosθ}{cosθ-μ_s sinθ}$ = $\frac{tanθ-μ_s}{1+μ_stannθ}$    

∴ $v_{min}=\sqrt{\frac{rg(tanθ-μ_s)}{1+μ_stannθ}}$………..(3)

For tanθ ≤  μ_s (as on most rough roads),

v_min = 0 (i.e., a car can be brought to a halt without sliding down).

(b) For maximum safe speed :If the car is driven fast enough, at a speed greater than the optimum speed v, it may skid off up the incline so that f_s is down the incline, see Fig.

Fig. Maximum safe speed on a banked road

Resolve N and f_s into two perpendicular components : N cosθ vertically up and f_s sinθ vertically clown; N sinθ and f_s cosθ horizontally towards the centre of the circular path. So long as the car takes the turn without skidding off, the horizontal components N sinθ and f_s cosθ together provide the necessary centripetal force, and N cosθ balances the sum mg + f_s sinθ. If v_max, is the maximum safe speed without skidding,

$\frac{v^2_{max}}{r}$= N sinθ + f_s cosθ = N(sinθ + μ_scosθ) ………..(4)

And N cosθ =mg + f_s sinθ = mg + μ_sNsinθ

∴  mg = N(cosθ - μ_ssinθ)  ………..(5)

Dividing Eq. (4) by Eq (5)

∴ $\frac{v^2_{max}}{rg}$ = $\frac{sinθ+μ_s cosθ}{cosθ-μ_s sinθ}$ = $\frac{tanθ+μ_s}{1-μ_stannθ}$    

∴ $v_{max}=\sqrt{\frac{rg(tanθ+μ_s)}{1-μ_stannθ}}$ ………..(6)

1- μ_stanθ = 0, i.e. μ_s = cotθ, v_max = ∞

Ignoring few special cases, the maximum value of μ_s  = 1.

Thus, for θ ≥ 45°, v_max = ∞

i.e., on a heavily banked road a car is unlikely to skid up the incline and the minimum limit is more important.

Expression for the angular speed of the bob of a conical pendulum :

 A conical pendulum is a small bob suspended from a string and set in UCM in a horizontal plane with the centre of its circular path below the point of suspension such that the string makes a constant angle θ with the vertical.

Consider a conical pendulum of string length L with its bob of mass m performing UCM along a circular path of radius r (Fig).

At every instant of its motion, the bob is acted upon by its weight mg and the tension T in the string.

If the constant angular speed of the bob is w, the necessary horizontal centripetal force is

T₀ = mω²r

T₀. is the resultant of the tension in the string and the weight. Resolve T into components T₀ cosθ vertically opposite to the weight of the bob and

T₀ sinθ horizontal. T₀ cosθ  balances the weight. T₀ sinθ is the necessary centripetal force.

∴ T₀ sinθ = mω²r  ……… (1)

and T₀ cosθ  = mg  ……… (2)

Dividing Eq. (1) by Eq. (2)

Tanθ = $\frac{ω^2}{r}$

From the diagram,

Tanθ = $\frac{r}{AC}$ = $\frac{r}{h}$ = $\frac{r}{Lcosθ}$    ……… (3)

∴ $\frac{r}{h}$ =  $\frac{ω^2r}{g}$

∴ ω² = $\frac{g}{h}$ = $\frac{g}{Lcosθ}$    ……… (4)

The angular speed of the bob,

ω = $\sqrt{\frac{g}{h}}$ = $\sqrt{\frac{g}{Lcosθ}}$   ……… (5)

From Eq. (4), cosθ=g/ω²L. Therefore, as ω increases, cosθ decreases and θ increases

Expression for the frequency of revolution of the bob of a conical pendulum :

If n is the frequency of revolution of the bob,

ω = 2πn = $\sqrt{\frac{g}{Lcosθ}}$

∴ n =  $\frac{1}{2π}\sqrt{\frac{g}{Lcosθ}}$  ..is the required expression for the frequency.

From the above expression, we can see that

(i) n ∝ $\sqrt{g}$    (g is the acceleration due to gravity at the place.)

(ii) n ∝ $\frac{1}{\sqrt{L}}$

(iii) n ∝ $\frac{1}{\sqrt{cosθ}}$     …..(if θ increases, cosθ decreases and n increases)

(iv) The frequency is independent of the mass of the bob.

Period of a conical pendulum and an expression for it :  The period of a conical pendulum is the time taken by its bob to complete one revolution in a horizontal circle with constant speed.

For the derivation, refer above expression up to Eq. (5) and continue.

If T_P is the period, ω = 2π/T_P

∴ T_P = 2π/ω = $2π\sqrt{\frac{Lcosθ}{g}}$  is the required expression.

Note - L cosθ =AC = h, where h is the axial height of the cone.

∴ T_P = $2π\sqrt{\frac{h}{g}}$

This shows that the period of a conical pendulum is the same as that of a simple pendulum of length h

From the above expression, we can see that

(i) Time period T_P ∝  $\sqrt{L}$

(ii) T_P ∝ $\frac{1}{\sqrt{g}}$

(iii) T_P ∝ $\sqrt{cosθ}$     …..(if θ increases, cosθ and T_P decreases)

(iv) The period is independent of the mass of the bob.