## Centripetal Force

We have seen that the value of normal or centripetal acceleration in uniform circular motion is \(\omega^2 r \text { or } \frac{v^2}{r}\). So, if the mass of a body is m, then effective force on the body

= mass x acceleration = \(m \omega^2 r=\frac{m v^2}{r}\)

- Since the acceleration is towards the centre, the effective force acting on the body will also be towards the centre. This force is known as centripetal force.
- According to Newton’s first law of motion, we know that if no external force is acting on a body, the body will remain at rest or will move with uniform linear velocity, i.e., the body will not move in a circular path.
- So, to move a body in a circular path some external force must be acted on it. During motion in a circular path, the acceleration of the body must be towards the centre.
- So, the force acting on the body must be towards the centre of the circular path, i.e., radially inwards. This external force is nothing but the centripetal force.

**Read and Learn More: Class 11 Physics Notes**

**Centripetal Force Definition:** Centripetal force is the force which moves the body in a circular path. It acts perpendicular to the direction of linear velocity and is directed radially towards the centre of that circular path.

If the mass of the body is m, the radius of the circular path is r, the linear velocity of the body is v and its angular velocity is ω, then centripetal force = \(m \omega^2 r=\frac{m v^2}{r}\).

Here, it is to be mentioned that the centripetal force is a no-work force. In the direction of centripetal force, no displacement of the body occurs, and hence, this force does not do any work.

**Centripetal Force Some Practical Examples:**

- A stone is rotated along a circular path by a string. The string pulls the stone towards the centre. This pull or tension is the centripetal force.
- While riding a bicycle along a circular path, the frictional force acting between the tyre of the cycle and the road supplies the centripetal force necessary to move the bicycle along that circular path.
- The gravitational attraction of the sun on any planet provides the planet with the necessary centripetal force to revolve around the sun.
- The electrostatic attraction between the positively charged nucleus inside an atom and the negatively charged electrons supplies the electrons with the necessary centripetal force to revolve around the nucleus.

## Some Practical Examples Of Centripetal Force

**Motion Of A Cyclist On A Horizontal Circular Track:** When a cyclist goes around a horizontal circular track, a centripetal force is required. The force of friction between the tyres and the road is too small to provide the necessary centripetal force.

- As a result, a cyclist going round a curve leans inward, because then the horizontal component of the normal reaction provides the necessary centripetal force.
- Let the angle with the vertical at which the cyclist leans be θ. In this situation, the vertical component of the reaction R offered by the ground on the cyclist = OB = Rcosθ and the horizontal component of R = OA = Rsinθ.
- The vertical component balances the weight of the cycle along with the cyclist and the horizontal component supplies the necessary centripetal force required to take the turn on the circular path.

If the weight of the cycle along with the cyclist is mg, the radius of the circular path is r and the velocity of the cyclist is v, then Rcosθ = mg …(1)

R \(\sin \theta=\frac{m v^2}{r}\) (necessary centripetal force)…(2)

Dividing equation (2) by equation (1) we get, \(\tan \theta=\frac{v^2}{r g}\)….(3)

From equation (3), it is evident that the more the velocity (v) of the cyclist or less the radius (r) of the circular path, the more the value of θ will be, i.e., the cyclist has to incline more towards the bend.

Since the normal force of the ground, Rcosθ = mg, the limiting frictional force between the tyre and the road is fμmg (μ = coefficient of friction). It is clear that this limiting frictional force is the maximum value of the centripetal force.

If the speed of the cyclist is v_{m} for a circular path of certain radius r, the centripetal force will be maximum when, \(\frac{m v_m^2}{r}=\mu m g \text { or, } v_m^2=\mu r g \text { or, } v_m=\sqrt{\mu r g}\)…(4)

which it can turn on a circular path of radius r. If the cyclist tries to attain a speed more than this maximum speed on that road, the frictional force cannot provide the necessary centripetal force, and hence, the cycle will skid. If the maximum value of the angle of inclination with the vertical is θ_{m}, then \(\tan \theta_m=\frac{v_m^2}{r g}=\frac{\mu r g}{r g}=\mu\)….(5)

It should be noted that none of the equations (3), (4) and (5) depend on the mass (m) of the cycle along with the cyclist.

## Centripetal Force Numerical Examples

**Example 1. A cyclist speeding at 18 km/h on a level rod takes a sharp circular turn of a radius 3 m without reducing the speed and without spending towards the centre of the circular path. The coefficient of static friction between the tyres and the road is 0.1. Will the cyclist slip while taking the turn?**

**Solution:**

The maximum safe speed is given by, \(v_{\max }=\sqrt{\mu \mathrm{rg}}=\sqrt{0.1 \times 3 \times 9.8}=1.715 \mathrm{~m} / \mathrm{s}\)

Actual speed of the cyclist, \(v=18 \mathrm{~km} / \mathrm{h}=\frac{18 \times 1000}{60 \times 60} \mathrm{~m} / \mathrm{s}=5 \mathrm{~m} / \mathrm{s}\)

Since the actual speed is greater than the maximum safe speed, the cyclist will slip while taking the turn.

**Example 2. A cyclist is moving in a circular path of radius 20 m with a velocity of 18 km ·h ^{-1}. What is his inclination toward the vertical?**

**Solution:**

Let the angle of inclination of the cyclist with the vertical be θ.

The magnitude of the velocity, i.e., speed of the cyclist, \(\nu=\frac{18 \times 1000}{60 \times 60}=5 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

and radius of the circular path ( r) = 20 m.

We know that, \(\tan \theta=\frac{v^2}{r g}=\frac{5 \times 5}{20 \times 9.8}=0.1276\)

∴ \(\theta=\tan ^{-1}(0.1276)=7.27^{\circ}\)

**Example 3. A mass is suspended from the ceiling by a string revolving in a horizontal circle of radius 5 cm. The tangential speed of the mass is 0.7 m · s ^{-1}. What is the angle between the string and the vertical? (Consider acceleration due to gravity as 9.8 m · s^{-2})**

**Solution:**

Let the angle between the string and the vertical be θ.

From the \(\tan \theta=\frac{\frac{m v^2}{r}}{m g}=\frac{v^2}{r g}\)

Here, \(r=5 \mathrm{~cm}=0.05 \mathrm{~m}\),

v = \(0.7 \mathrm{~m} \cdot \mathrm{s}^{-1}, g=9.8 \mathrm{~m} \cdot \mathrm{s}^{-2}\)

So, \(\tan \theta=\frac{(0.7)^2}{0.05 \times 9.8}=1 \quad or, \theta=45^{\circ}\).

**Example 4. A coin is placed on a horizontal turntable rotating at 33 1/3 rpm. The coin revolves with the table without slipping provided the coin is not more than 10 cm away from the axis. How far from the axis can the coin be placed so that it revolves with the table without slipping if the turntable rotates at 45 rpm? (g = 980 cm · s ^{-2})**

**Solution:**

The magnitude of the limiting frictional force between the coin and the table is the maximum value of centripetal force. So, if the coin of mass m is rotating along the turntable in a circular path of maximum radius r with angular velocity co, limiting frictional force,

f = mω²r

In the first case of the given problem, \(f=m \omega_1^2 r_1\)

and in the second case, \(f=m \omega_2^2 r_2\)

∴ \(m \omega_1^2 r_1=m \omega_2^2 r_2\)

or, \(r_2=\frac{\omega_1^2}{\omega_2^2} \times r_1=\left(\frac{100}{3 \times 45}\right)^2 \times 10=5.5 \mathrm{~cm}\).

**Example 5. A small body is kept at a distance of 7 cm from the centre of a gramophone disc. The disc starts rotating with gradually increasing speed and the body is just on the verge of being thrown off when the disc rotates at 60 rpm. What will be the rate of rotation of the disc when the body, kept at a distance of 12 cm from the centre, is just being thrown off?**

**Solution:**

The magnitude of the limiting frictional force between the body and the disc gives the maximum value of the centripetal force. So, if the body of mass m rotates along a circular path of maximum radius r with an angular velocity, then the limiting frictional force, f = mω²r.

In the given problem, f = \(m \omega_1^2 r_1\), in the first case

and \(f=m \omega_2^2 r_2\), in the second case

∴ \(m \omega_1^2 r_1=m \omega_2^2 r_2 \quad or, \omega_2^2=\omega_1^2 \cdot \frac{r_1}{r_2}\)

or, \(\omega_2=\omega_1 \sqrt{\frac{r_1}{r_2}}=60 \times \sqrt{\frac{7}{12}}=45.8 \mathrm{rpm}\).

**Example 6. The driver of a truck travelling with a velocity v suddenly notices a wall in front of him at a distance d. Is it better for him to apply brakes or to make a circular turn without applying brakes in order to just i avoid crashing into the wall?**

**Solution:**

To stop the truck by applying brakes, the final velocity should be zero.

So, applying the equation v² = u² – 2 as we get, 0 = v² -2ad

or, a = \(\frac{v^2}{2 d}\) = retardation of the truck.

If the mass of the truck is m, then the force applied by the brakes, F = ma = \(\frac{m v^2}{2 d}\)

- If the truck turns along a circular path to avoid collision, without applying brakes, then the radius of that circular path = d. For this circular motion, the necessary centripetal force is, \(F^{\prime}=\frac{m v^2}{d}\)
- Here, F< F’; hence, it is better to stop the truck by applying brakes so that crashing can be avoided by the application of comparatively less force.
- Here, it should be noted that due to the frictional force exerted by the road on the lyre of the wheels of the truck, it is possible to stop the truck.
- After the application of brakes, if the road does not exert any frictional force on the tyre then (1) due to the application of frictional force by the brakes on the wheels, they stop rotating and (2) the wheels go on skidding on the road.

(There is a chance of skidding of wheels on wet road when brakes are applied in a fast-moving vehicle). Hence, for any vehicle, if the frictional force necessary to stop it is less, then the chance of skidding is lowered.

**Example 7. A body is kept at rest at a distance of 10 cm from the centre of a gramophone disc. If the coefficient of friction between the body and the disc is 0.3, then for what maximum rps of the disc, will the body not be thrown off the disc?**

**Solution:**

The magnitude of the limiting frictional force between the body and the gramophone disc is the maximum value of the centripetal force. If the mass of the body is m and it rotates with the disc along a circular path of radius r with angular velocity co such that the body will not be thrown off the disc, then

limiting frictional force, f = mω²r

or, μmg = mω²r [ω = 2πn, n = number of revolutions of the disc per second (rps); μ = coefficient of friction between the body and the disc]

or, \(\mu g=(2 \pi n)^2 r=4 \pi^2 n^2 r\) or, \(\quad n^2=\frac{\mu g}{4 \pi^2 r}\)

or, \(n=\sqrt{\frac{\mu g}{4 \pi^2 r}}=\sqrt{\frac{0.3 \times 980}{4 \times \pi^2 \times 10}}\)

= \(0.863 \mathrm{rps}\)

**Motion Of A Train Or A Car On A Circular Track**

**Motion Along A Horizontal Path:** Let O be the point about which a train or a car takes a turn in a circular path while moving on a horizontal path. The effective centripetal force acting on the car is \(\frac{m v^2}{r}\), where m = mass of the car, v = speed of the car, r = OA = radius of the circular path.

Since the frictional force f between the wheel and the road supplies the necessary centripetal force, f = \(\frac{m v^2}{r}\)…(1)

Again the normal force R on the wheel by the ground is equal to the weight of the car, i.e., R = mg….(2)

**Condition For No Skidding:** If the frictional force f between the wheels and the road becomes equal to the limiting friction and if the coefficient of friction is f then f = μR

Using the equations (1) and (2) we get, \(\frac{m v^2}{r}\) = μmg or, v² = μrg

So, the maximum velocity of the car with which it can turn on a circular path of radius r will be \(v_m=\sqrt{\mu r g}\)…(3)

**Banking In A Circular Path:** A moving car, while taking a turn, requires a centripetal force. At the turning point if the road is banked, i.e., the road does not remain horizontal, then the horizontal component of the normal force offered by the road on the moving car supplies the necessary centripetal force.

- If this component is not sufficient to provide the necessary centripetal force, then the remaining part of it comes from the horizontal component of the frictional force between the road and the car.
- So, if the road is properly banked, then the horizontal component of the reaction force alone can provide the necessary centripetal force. In that case, frictional force does not come into play.
- To avoid accidents of cars near a bend, the centripetal force is supplied by raising the outer edge of the road above its inner edge and this method of construction of a road is known as banking.
- The angle made by the road with the horizontal is known as the angle of banking.

The angle of banking of the road ∠BAC = θ, speed of the car = v, mass of the car = m and the radius of the bend of the road = r (i.e., the radius of the circular path of the car).

**In the above-mentioned case, there are two forces acting on the car:**

- Weight of the car mg acting vertically downwards and
- Reaction force R offered by the road on the car, normal to the surface of the road.

The horizontal component Rsinθ of R provides the necessary centripetal force and the vertical component Rcos8 balances the weight of the car.

So, \(R \sin \theta=\frac{m v^2}{r}, R \cos \theta=m g\)

∴ \(\tan \theta=\frac{v^2}{r g}\)…(4)

From the above relation, the angle of banking (θ) required for a car of certain speed (v) moving on a particular road can be ascertained. For a particular speed, the value of θ is fixed, i.e., a certain road is properly banked only for a definite value of the speed.

Moreover, according to the figure, tan0 = \(\frac{h}{x}\)

(h = the height of the outer edge of the road with respect to its inner edge; it should be noted that the inner edge of the road lies towards the centre of the circular path)

∴ \(\frac{v^2}{r g}=\frac{h}{x}\) (x = horizontal distance between the outer and the inner edges of the road)

or, \(v=\sqrt{\frac{h r g}{x}}\)…..(5)

- This relation expresses the maximum speed of the car with which it moves along a banked circular path. If the speed is more than this, then there will be a chance of the car skidding outwards.
- Similarly, railway tracks are also banked near a bend. Here, the outer rail near the bend is placed at a slightly higher level than the inner rail.
- At the site of every bend, the maximum permissible speed is displayed on a board. To avoid an accident, the driver of the train should be acquainted with the maximum permissible speed at that place.

**Condition For No Skidding:** If the speed of the car exceeds the maximum permissible speed, then the car tends to skid towards point B and the frictional force f acting along the direction BA tries to oppose this skidding.

Let the maximum permissible speed of the car for limiting friction is v_{m}. Then limiting frictional force is f = μR (μ = coefficient of friction between the wheel and the road).

According to the given figure, the condition for no skidding is

R \(\sin \theta+f \cos \theta=\frac{m v_m^2}{r}\) or, \(R \sin \theta+\mu R \cos \theta=\frac{m v_m^2}{r}\)

or, \(R(\sin \theta+\mu \cos \theta)\) = \(\frac{m v_m^2}{r}\)….(6)

Again, \(R \cos \theta=m g+f \sin \theta\) or, \(R \cos \theta-\mu R \sin \theta=m g\)

or, \(R(\cos \theta-\mu \sin \theta)=m g\)….(7)

(6)+(7) gives, \(\frac{\nu_m^2}{r g}=\frac{\sin \theta+\mu \cos \theta}{\cos \theta-\mu \sin \theta}=\frac{\tan \theta+\mu}{1-\mu \tan \theta}\)

or, \(v_m=\left[r g\left(\frac{\mu+\tan \theta}{1-\mu \tan \theta}\right)\right]^{1 / 2}\)….(8)

So, when the speed of the car exceeds v_{m}, it skids towards the point B. Naturally, for θ = 0, i.e., on an unbanked road, \(v_m=\sqrt{\mu r g}\) which is nothing but equation (3).

In equation (8) the quantity (1 – μtanθ) is less than 1, and hence, the value of v_{m} becomes greater due to the presence of banking. Hence, a car can take a sharp turn on a banked road with speed than on an unbanked horizontal road.

## Circular Motion

## Circular Track Numerical examples

**Example 1. The radius of curvature of a railway track at a bend is 500 m. The distance between the two tracks is 1 m and the outer line is 4 cm higher than the inner. At what maximum speed can a train bend around the curve without exerting lateral pressure on the outer line?**

**Solution:**

If the speed of the train is v, the radius of curvature of the tracks is r and the angle of banking is θ, then \(\tan \theta=\frac{v^2}{r g}\)

If the distance between the two lines is x and the difference in their heights is h, then \(\tan \theta=\frac{h}{x}\)

∴ \(\frac{v^2}{r g}=\frac{h}{x} \text { or, } v^2=\frac{h r g}{x} \quad \text { or, } \quad v=\sqrt{\frac{h r g}{x}}\)

∴ v = \(\sqrt{\frac{0.04 \times 500 \times 9.8}{1}}=14 \mathrm{~m} \cdot \mathrm{s}^{-1}=50.4 \mathrm{~km} \cdot \mathrm{h}^{-1}\)