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Q6.16:
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 :

Step 1: Identification of Given Parameters

Let $M$ be the mass of the metre stick and $L = 100 \text{ cm}$ be its total length. The centre of mass of a uniform metre stick is located at the $50.0 \text{ cm}$ mark.

  • Mass of metre stick = $M$
  • Position of metre stick center of gravity ($x_g$) = $50.0 \text{ cm}$
  • Mass of two coins ($m$) = $2 \times 5 \text{ g} = 10 \text{ g} = 0.01 \text{ kg}$
  • Position of coins ($x_c$) = $12.0 \text{ cm}$
  • New fulcrum position (knife edge) ($x_f$) = $45.0 \text{ cm}$

Step 2: Defining the Principle of Moments

For the metre stick to be in rotational equilibrium at the new fulcrum, the clockwise torque must equal the counter-clockwise torque [Per the Principle of Moments: $\sum \tau = 0$].

The torque $\tau$ exerted by a force $F$ at a distance $d$ from the pivot is given by $\tau = F \cdot d$.

Step 3: Calculation of Lever Arms

  • Distance of the metre stick's center of mass from the fulcrum ($d_1$): $d_1 = |x_g - x_f| = |50.0 \text{ cm} - 45.0 \text{ cm}| = 5.0 \text{ cm} = 0.05 \text{ m}$
  • Distance of the coins from the fulcrum ($d_2$): $d_2 = |x_c - x_f| = |12.0 \text{ cm} - 45.0 \text{ cm}| = |-33.0 \text{ cm}| = 33.0 \text{ cm} = 0.33 \text{ m}$

Step 4: Algebraic Formulation and Solving for M

Equating the moments about the fulcrum at $45.0 \text{ cm}$:

$M \cdot g \cdot d_1 = m \cdot g \cdot d_2$ [Where $g$ is the acceleration due to gravity]

$M \cdot 0.05 = 0.01 \cdot 0.33$

$M = \frac{0.01 \times 0.33}{0.05}$

$M = \frac{0.0033}{0.05} = 0.066 \text{ kg}$

$M = 66 \text{ g}$

Visual Representation of the Lever System

Coins Fulcrum (45cm) CoM (50cm)

Final Solution: The mass of the metre stick is 66 g.


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