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SS1: PHYSICS – 1ST TERM

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  1. Introduction to Physics | Week 1
    4Topics
    |
    1 Quiz
  2. Measurement | Week 2
    3Topics
  3. Measurement of Mass | Week 3
    6Topics
    |
    1 Quiz
  4. Motion | Week 4
    5Topics
    |
    1 Quiz
  5. Velocity-Time Graph | Week 5
    4Topics
    |
    1 Quiz
  6. Causes of Motion | Week 6
    5Topics
    |
    1 Quiz
  7. Work, Energy & Power | Week 7
    3Topics
  8. Energy Transformation / Power | Week 8
    3Topics
    |
    1 Quiz
  9. Heat Energy | Week 9
    5Topics
    |
    1 Quiz
  10. Linear Expansion | Week 10
    6Topics
    |
    1 Quiz
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Circular motion is a movement of an object along the circumference of a circle or rotation along a circular path. It can be uniform with a constant angular rate of rotation and constant speed or non-uniform with a changing rate of rotation.

For example, an artificial satellite orbiting the earth at a constant height, a ceiling fan’s blade rotating around a hub, a stone which is tied to a rope and whirled in circles, a car turning through a curve in a race track, planets moving around the sun.

centripetal acceleration

If acceleration is the time rate of change of velocity, a change in the direction of velocity also constitutes a change in magnitude of the acceleration.

The acceleration is always directed towards the centre of the circular path and this is called centripetal acceleration.

a =  \( \frac {v^2}{r}\)

v = Velocity of the object, and it is always directed along the tangent to the circle at any point of the path.

r = radius of the circle

The inwardly directed acceleration leads to a force known as centripetal force, which is also directed towards the centre of the circle.

Centripetal force FT is defined as that inward force required to keep an object moving with a constant speed in a circular path.

Centripetal Force = Mass x Acceleration

= \(\scriptsize M \times \normalsize \frac {V^2}{r} \)   (m = mass of the object)

FT = \(\frac {MV^2}{r} \)

As the inward force keeps trying to keep the object moving in a circular path, there is also an outward force trying to pull the object out of the circular path and this is called centrifugal force.

Centrifugal force is a balancing force that counterbalances the centripetal force, so as to prevent the Centripetal force from pushing the object to the centre of the circle.

Centrifugal force is opposite to Centripetal force.

centrifugal force

Since centrifugal force is a balancing force, it is equal in magnitude but opposite in direction to the centripetal force.

Therefore,

The formula for both centrifugal and centripetal force is the same:

Centripetal force = Centrifugal force, F = \( \frac {MV^2}{r} \)

Example 1

A mass of 15 kg is moving in a circular path of radius 2.5m with a uniform speed of 45 \( \scriptsize ms^{-1} \). Find the centripetal acceleration and the corresponding centripetal force.

Solution

Data given:

mass = 15kg

radius = 2.5m

uniform speed = 45 \( \scriptsize ms^{-1} \)

Cetripetal acceleration is given by

a = \( \frac{v^2}{r} \\ = \frac{45 \: \times \: 45}{2.5} \\ = \scriptsize 1012.5 ms^{-2} \)

Centripetal force is given by

F = mass x acceleration

= 15kg x 1012.5 ms-2

= 15187.5 kgms-2

= 15187.5 N

Example 2

Calculate the magnitude of centripetal force on a car of mass 6.5 kg that is moving round a circle of radius 2m with a a velocity of 10m/s.

Also, determine the centripetal acceleration of the object.

Solution

Data given: mass =  6.5 kg, radius = 2m, velocity = 10 m/s

Use the formula,

F = \( \frac {MV^2}{r} \\ = \frac{6.5 \: \times \: 10^2}{2} \\ = \frac{6.5 \: \times \: 100}{2} \\ = \scriptsize 325 N \)

Centripetal acceleration of the car, a = \( \frac{V^2}{r} \)

= \( \frac{10^2}{2} \\ = \frac{100}{2} \\ \scriptsize = 50 ms^{-2} \)

Example 3

If 500 N force acts on a train of mass 45 kg, moving round a circle with a velocity V, of radius 2.70 m, calculate the velocity of the train and hence its centripetal acceleration?

Solution

Data Given: force = 500N, mass = 45kg, velocity, V = ?, raius = 2.70m

Use the formula,

F = \( \frac {MV^2}{r} \)

We are looking for the velocity, so let’s make v the subject of the formula.

\( \frac{MV^2}{r} = \scriptsize F \)

multiply both sides by r

\( \frac{MV^2}{r} \scriptsize \: \times \: r = \scriptsize F \: \times \: r\)

\( \scriptsize MV^2 = Fr \)

Divide both sides by M

\( \frac{MV^2}{M} = \frac{Fr}{M} \)

\( \scriptsize V^2 = \normalsize \frac{Fr}{M} \)

Find the square root of both sides

\( \scriptsize \sqrt{V^2} = \normalsize \sqrt{\frac{Fr}{M}} \)

\( \scriptsize V = \normalsize \sqrt{\frac{Fr}{M}} \)

Now let’s substitute in the data given.

\( \scriptsize V = \normalsize \sqrt{\frac{500 \: \times \: 2.7}{45}} \)

\( \scriptsize V = \normalsize \sqrt{\frac{500 \: \times \: 2.7}{45}} \\ = \sqrt{ \frac{1350}{45}} \\ = \scriptsize \sqrt{30} ms^{-1}\)

V = \( \scriptsize 5.477 ms^{-1} \)

Centripetal acceleration of the car, a = \( \frac{V^2}{r} \)

= \( \frac{5.477^2}{2.70} \\ = \frac{30}{2.70} \\ \scriptsize = 11.11 ms^{-2} \)

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