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Standard XII

Physics

Displacement of COM:Application

A meter stick...

Question

A meter 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 $12.0 c m$ mark, the stick is found to be balanced at $45.0 c m$ . What is the mass of the meter stick?

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Solution

Figure shows the forces on the stick
The mass of metre stick is concentrated at $C M = 50 c m$
Consider the torques around $0$ .Since the stick is in rotational equilibrium,
${¯ τ}_{e x t} = m (5 c m) - (10 g) (45 - 12) = 0$
$∴= 10 g \frac{(33 c m)}{(5 c m)} = 66 g m .$

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A meter stick is balanced on a knife edge at its centre. When two coins each of mass 5g are put one on top of the other at 12.0cm mark, the stick is found to be balanced at 45.0cm. What is the mass of the meter stick?

Figure shows the forces on the stickThe mass of metre stick is concentrated at CM=50cmConsider the torques around 0.Since the stick is in rotational equilibrium,¯τext=m(5cm)−(10g)(45−12)=0∴=10g(33cm)(5cm)=66gm.

A meter 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 $12.0 c m$ mark, the stick is found to be balanced at $45.0 c m$ . What is the mass of the meter stick?

Figure shows the forces on the stick
The mass of metre stick is concentrated at $C M = 50 c m$
Consider the torques around $0$ .Since the stick is in rotational equilibrium,
${¯ τ}_{e x t} = m (5 c m) - (10 g) (45 - 12) = 0$
$∴= 10 g \frac{(33 c m)}{(5 c m)} = 66 g m .$