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SS2: PHYSICS - 1ST TERM

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  1. Scalars & Vectors | Week 1
    5 Topics
    |
    1 Quiz
  2. Equations of Motion | Week 2
    3 Topics
    |
    1 Quiz
  3. Projectile | Week 3
    5 Topics
  4. Equilibrium of Forces I | Week 4
    4 Topics
  5. Equilibrium of Forces II | Week 5
    4 Topics
  6. Stability of a Body | Week 6
    4 Topics
    |
    1 Quiz
  7. Simple Harmonic Motion (SHM) | Week 7
    4 Topics
  8. Speed, Velocity & Acceleration & Energy of Simple Harmonic Motion | Week 8
    5 Topics
    |
    1 Quiz
  9. Linear Momentum | Week 9
    6 Topics
    |
    1 Quiz
  10. Mechanical Energy & Machines | Week 10
    2 Topics
    |
    1 Quiz



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The Centre of gravity of a body is the point at which the total weight of a body appears to be concentrated.

or

The Centre of gravity of a body is the point at which the total weight of a body appears to pass through.

Centre of mass of a body is the point at which the total mass of a body appears to pass through or concentrated.

The Centre of gravity for objects varies as shown.

C.G

For a Uniform rod, the centre of gravity is at the mid-point of the centre rule.

Screenshot 2022 07 06 at 00.29.33

For a Uniform Circular Plate, the centre of gravity is at the mid-point of the circular plate

Screenshot 2022 07 06 at 00.34.30

For a Uniform Circular Ring, the centre of gravity is at the centre of the ring.

Screenshot 2022 07 06 at 00.37.11

For a Uniform Rectangular Sheet, the centre of gravity is at the point of intersection of the diagonals.

Screenshot 2022 07 06 at 00.45.30

For a Uniform Triangular Plate, the centre of gravity is at the point of intersection of the medians.

Determine the Position of the Centre of Gravity of an Irregular Object:

The centre of gravity of irregular-shaped objects can be determined by suspension with a plumb line from two or more points or by balancing the body on a knife-edge.

Use of a Plumbline:

Here we make use of a plumbline to find the centre of gravity of an irregularly shaped lamina.

A plumbline is a thread with a small heavy weight tied to one end.

plumb line
Plumbline.

Procedure:

  • Make a plumb line by tying a nut at one end of the string and nail or any other weight at the other end.
  • Make two holes A and B near the edge of the irregularly shaped lamina.
Lamina 2
  • Pin the lamina through hole A so that it is able to swing freely from horizontal support. A needle or nail can be used.
  • Suspend a plumbline from the pin.
centre of gravity of a lamina
  • Once the setup becomes steady, draw a vertical line through A along the thread on the cardboard with the help of a rule or a straight edge.
  • Repeat the procedure from step 3 with hole B.
  • The point of intersection of the two lines gives the centre of gravity of the lamina.
C.G 3
The intersection of the two plumblines is the object’s centre of gravity.

Using the Balancing Method:

A solid body can be balanced by supporting it on a knife-edge at its centre of gravity. For example, a uniform metre rule has its centre of gravity at the mark 50cm. It is possible because the algebraic sum of moments of the weights of all particles of the rule about any point is zero. This point where the rule is supported is its centre of gravity.

Determining the Weight of an Object by the Principle of Moments:

The weight of an object, e.g a metre rule, can be determined using the principle of moments. This is shown in the example below.

Example:

Consider the diagram below of a uniform bar that is 1m long, hanging on a knife-edge at the 30cm mark and in equilibrium under the action of different forces.

Determine;
(i) Total clockwise moments;
(ii) Total anticlockwise moments;
(iii) The weight of the bar;

eg4

Solution:

First, we need to label the centre of gravity, which is where the weight of the bar is acting. This point will be the midpoint of the bar.

The length of the bar is 1 m or 100 cm, therefore, the weight is acting at the centre which is the 50cm mark.

The distance between 2N force and centre of gravity and fulcrum and centre of gravity can be calculated as shown below.

eg5

(i) Total clockwise moments = 2N x (30 + 20) + W x 20

= (100 + 20W)Ncm

(ii) Total anticlockwise moments = 15 x 30 = 450Ncm

(iii) The weight of the bar;

The weight of the bar = W

At equilibrium

Total Clockwise moments = Total Anticlockwise moments

100 + 20W = 450

20W = 450 – 100

20W = 350

W = \(\frac{350}{20}\)

W = 17.5N

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