More Solved examples involving Torque and Center of gravity

This page shows two more solved examples in continuation of the examples that we saw here: Solved examples involving torque.

Example 1
From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body
Solution:
1. The uniform circular disc can be supported at it's center as shown in fig.1(a) below:

Fig.1

• But when a hole is cut on the right side, the disc will tilt towards the left and fall off. This is shown in fig.1(b)
2. If we shift the support towards the left (along the diameter), the disc will balance again
• This is shown in fig.2(a) below:

Fig.2

• In fig.2(a), the disc is supported at the new CG
• We have to find the position of the new CG
3. In fig.2(b), the small circle is painted red and glued back to get the original disc
• The CG of this 'composite disc' will be same as the center of the original disc 
4. In such a situation, we apply the conditions of equilibrium
• Let the brown portion in fig.2(b) be indicated as 'Brown'
• Let the red portion in fig.2(b) be indicated as 'Red'
• Let 'mass per unit area' of the disc be M
• Since the disc is uniform, this 'M' will be same at all points on the disc
5. Fig.3(a) below shows the dimensions

Fig.3

• Area of Red = $\mathbf\small{\pi\,\left(\frac{R}{2}\right)^2=\frac{\pi R^2}{4}}$
    ♦ So $\mathbf\small{W_{Red}=\frac{\pi R^2Mg}{4}}$
• Area of Brown = $\mathbf\small{\pi\,R^2-\frac{\pi R^2}{4}=\frac{3\pi R^2}{4}}$
    ♦ So $\mathbf\small{W_{Brown}=\frac{3\pi R^2Mg}{4}}$ 
6. Fig.3(b) shows the forces
• The distance between the required CG and the support is denoted as 'x'
7. Since the composite disc is in translational equilibrium, we have:
R - W_Brown - W_Red = 0
8. Since the composite disc is in rotational equilibrium, we have: net torque = 0
• Let us take the torques about the support 
(i) Torque created by W_Brown about the support = W_Brown × x (anti clockwise)
(ii) Torque created by R about the support = 0
(iii) Torque created by W_Red about the support = W_Red × ^R⁄₂ (clockwise)
9. So applying the condition, we get:
-(W_Brown × x) + (W_Red × ^R⁄₂) = 0
$\mathbf\small{\Rightarrow x=\frac{R\,W_{Red}}{2\,W_{Brown} }}$
• Substituting the known values, we get: $\mathbf\small{x=\frac{\pi R^3Mg}{4}\div \frac{6\pi R^2Mg}{4}}$
$\mathbf\small{\Rightarrow x=\frac{\pi R^3Mg}{4}\times \frac{4}{6\pi R^2Mg}=\frac{R}{6}}$

Example 2
A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick?
Solution:
1. Given that, the metre stick is balanced on a knife edge at its centre 
• That means, if the support is given at the exact center, the metre stick will balance
• That means, it is a uniform scale
• This is shown in fig.2(a) below:

Fig.2

2. The condition when the two coins are placed at the 12 cm mark is shown in fig.b
• The forces and dimensions are also shown
• We see that, for equilibrium, the support has to be shifted by 5 cm towards the left
3. Since the system in fig.b is in translational equilibrium, we have:
R - W_Coin - W_Stick = 0
8. Since the system is in rotational equilibrium, we have: net torque = 0
• Let us take the torques about the support 
(i) Torque created by W_Coin about the support = W_Coin × 0.33 (anti clockwise)
(ii) Torque created by R about the support = 0
(iii) Torque created by W_Stick about the support = W_Stick × 0.05 (clockwise)
9. So applying the condition, we get:
-(W_Coin × 0.33) + (W_Stick × 0.05) = 0
W_Stick = (W_Coin × ^0.33⁄_0.05 ) = (10 ×  ¹⁄₁₀₀₀ ×  ^0.33⁄_0.05 ) = 0.066 kg = 66 g

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