Question

A flywheel used to prepare earthenware pots is set into rotation at 50 rpm. It is in the form of a disc of mass 5 kg and radius 0.2 m. A lump of clay (to be taken equivalent to a particle) of mass 0.8 kg falls on it and adheres to it at a certain distance 'x' from the centre. The value of 'x' if the wheel now rotates at 40 rpm, is

• A. \(1\) m
• B. \(\frac{1}{4}\) m
• C. \(\frac{1}{\sqrt{32}}\) m
• D. \(\frac{1}{\sqrt{8}}\) m

Hint:

No external torque acts, so angular momentum is conserved. The moment of inertia increases by the clay's contribution.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
No external torque acts, so \(I_{1}\omega_{1}=I_{2}\omega_{2}\), and for rpm values we can use the speeds directly.

Step 2: Calculation
Disc: \(I_{1}=\frac{1}{2}(5)(0.2)^{2}=0.1\ \text{kg m}^{2}\)
After clay sticks: \(I_{2}=0.1+0.8x^{2}\)
\(0.1\times50=(0.1+0.8x^{2})\times40\)

Step 3: Solve for x
\(0.125=0.1+0.8x^{2}\)
\(x^{2}=\dfrac{0.025}{0.8}=\dfrac{1}{32}\)
\(x=\dfrac{1}{\sqrt{32}}\ \text{m}\)

Final Answer:
The clay lands at a distance of \(\frac{1}{\sqrt{32}}\) m from the centre. \[ \boxed{\text{(C) }\dfrac{1}{\sqrt{32}}\ \text{m}} \]