Uniform Circular Motion (UCM) - Centripetal Acceleration

Circular motion is a type of motion where an object moves along a circular path. When you tie a stone to a string and rotate it, the stone moves in a circle. For this motion to happen, a force must act on the object continuously. This force is always directed towards the centre of the circle. Without this force, the object would move in a straight line. This special type of force is called centripetal force, which means "centre-seeking."

A force that acts on any object moving along a circle and is directed towards the centre of the circle. When this force stops acting, the object flies off along a straight line (tangent to the circle) in the direction of its velocity at that instant.

F = \[m\omega^{2}r=\frac{mv^{2}}{r}\] = mωv

where:

• F = Centripetal force (in Newtons)
• m = Mass of the object (in kg)
• ω = Angular speed (in rad/s)
• r = Radius of the circular path (in m)
• v = Linear speed or tangential velocity (in m/s)

a = ω²r = \[\frac {v^2}{r}\] = ωv

where:

• a = Centripetal acceleration (in m/s²)

• The centripetal force is always directed towards the centre of the circle
• It is perpendicular to the velocity of the object
• The velocity is always tangential to the circular path
• The magnitude of velocity remains constant in uniform circular motion, but its direction continuously changes
• The centripetal force is not a separate external force; it is the resultant of the actual forces acting on the object
• If the centripetal force stops acting, the object moves in a straight line along the tangent to the circle

For a particle moving in uniform circular motion with radius r and angular speed ω:

Position Vector:
\[\vec{r}=(r\cos[\omega t])\hat{i}+(r\sin[\omega t])\hat{j}\]

Velocity Vector (first derivative of position):
\[\vec{v}=\frac{d\vec{r}}{dt}=r\omega(-\sin[\omega t]\hat{i}+\cos[\omega t]\hat{j})\]

Acceleration Vector (derivative of velocity):
\[\vec{a}=\frac{d\vec{v}}{dt}=-\omega^2(r\cos[\omega t]\hat{i}+r\sin[\omega t]\hat{j})=-\omega^2\vec{r}\]

The negative sign shows that the acceleration is opposite to the position vector, meaning it points towards the centre of the circle.

Magnitude of Centripetal Acceleration:
a = ω²r = \[\frac {v^2}{r}\] = ωv

Centripetal Force (using Newton's Second Law):
\[\vec{F}=m\vec{a}=-m\omega^2\vec{r}\]

Magnitude of Centripetal Force:
\[F=m\omega^2r=\frac{mv^2}{r}=m\omega v\]

The Moon revolves around the Earth in a definite orbit. During this motion:

• The direction of the Moon's motion constantly changes
• The speed of the Moon remains constant, but its velocity direction changes
• The Earth exerts a gravitational force on the Moon that is directed towards the Earth's centre
• This gravitational force acts as the centripetal force for the Moon's circular motion
• Similarly, the Sun attracts all planets, including Earth, towards itself, providing the centripetal force for their orbital motion

• Centripetal force is essential for any circular motion to occur
• It explains why objects don't fly off when moving in circles
• Understanding centripetal force helps explain planetary motions and satellite orbits
• It is used in designing circular tracks for vehicles (banking of roads)
• Important in understanding the motion of electrons around the nucleus in atoms
• Critical in understanding the dynamics of rotating machinery and equipment
• Helps explain natural phenomena like the motion of celestial bodies

Aim: 

To demonstrate that a centripetal force directed towards the centre is required to keep an object moving in a circular path.

Requirements:

• A stone or a small weight
• A string
• Your hand (to hold and rotate the string)
• A circular space to perform the rotation
  

Procedure

1. Tie the stone to one end of the string
2. Hold the other end of the string in your hand
3. Rotate the string so that the stone moves along a circular path
4. Observe the direction in which you are pulling the string
5. Continue rotating the stone in a steady circular motion
6. Suddenly release the string.

A stone tied to a string, moving along a circular path, and its velocity in the tangential direction

Observation

• As you rotate the string, you continuously pull the stone towards the centre of the circle
• The stone moves in a circular path as long as you hold and rotate the string
• When you release the string, the stone immediately flies off
• The stone does not continue moving in a circle; instead, it moves in a straight line
• The direction of the stone's straight-line motion is tangent to the circle at the point of release

Conclusion:

A force is acting on the stone continuously, directed towards the centre of the circle. This force is called the centripetal force. When the centripetal force stops acting (the string is released), the stone moves in a straight line (tangent to the circle) because that is the direction of its velocity at that instant. This proves that centripetal force is necessary to maintain circular motion

Problem: An object of mass 50 g moves uniformly along a circular orbit with an angular speed of 5 rad/s. If the linear speed of the particle is 25 m/s, what is the radius of the circle? Calculate the centripetal force acting on the particle.

Solution:

• Mass (m) = 50 g = 0.05 kg
• Angular speed (ω) = 5 rad/s
• Linear speed (v) = 25 m/s
• Radius (r) =?
• Centripetal force (F) = ?

Find the radius using the relationship between linear and angular speed
v = ωr
r = \[\frac {v}{ω}\] = \[\frac {25}{5}\] = 5 m

Calculate centripetal force using the formula
F = \[\frac{mv^{2}}{r}\]
F = \[\frac{0.05\times25\times25}{5}\] = 6.25 N

Answer: The radius of the circle is 5 m, and the centripetal force acting on the particle is 6.25 N.

Problem: An object is travelling in a horizontal circle with uniform speed. At t = 0, the velocity is \[\vec u\] = 20\[\hat i\] + 35\[\hat j\] km/s. After one minute the velocity becomes \[\vec v\] = −20\[\hat i\] − 35\[\hat j\] km/s. What is the magnitude of the acceleration?

Solution:

Find the magnitude of initial velocity
u = \[\sqrt{(20)^2+(35)^2}=\sqrt{400+1225}=\sqrt{1625}\] = 40.3 m/s

Observe that the velocity has reversed direction
Since the velocity has changed from \[\vec u\] to −\[\vec u\], the object has completed half a circle.

Determine the time period of revolution
If the object completes half a circle in 1 minute, then the full period T = 2 minutes = 120 seconds.

Find the radius using the relationship
T = \[\frac {2πr}{u}\]
r = \[\frac {uT}{2π}\]

Calculate centripetal acceleration
a = \[\frac{u^2}{r}=\frac{u^2\times2\pi}{uT}=\frac{2\pi u}{T}\]
a = \[\frac{2\times3.14\times40.3}{120}\] = 2.11 m/s²

Answer: The magnitude of the acceleration is 2.11 m/s².

• Vehicles on Circular Tracks: When a car moves on a circular road, friction between the tyres and the road provides the centripetal force to keep the car moving in a circle
• Planetary Motion: All planets, including Earth, revolve around the Sun. The gravitational force from the Sun provides the centripetal force
• Moon's Orbit: The Moon revolves around Earth due to Earth's gravitational force acting as a centripetal force
• Spinning Coin: A coin on a rotating circular disk flies off tangentially when the disk rotates fast enough because the friction force (centripetal force) becomes insufficient
• Merry-Go-Round: When you sit on a rotating merry-go-round, the seat exerts a force towards the centre that keeps you moving in a circle
• Electron Motion: Electrons moving around the nucleus experience centripetal force due to the electric attraction of the nucleus