MCQ

A particle is rotated in a vertical circle by connecting it to a string of length l and keeping the other end of the string fixed. The minimum speed of the particle when the string is horizontal for which the particle will complete the circle is

#### Options

- \[\sqrt{gl}\]
- \[\sqrt{2gl}\]
- \[\sqrt{3gl}\]
- \[\sqrt{5gl}\]

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#### Solution

\[\sqrt{3gl}\]

Suppose that one end of an extensible string is attached to a mass m, while the other end is fixed. The mass moves with a velocity v in a vertical circle of radius R. At some instant, the string makes an angle θ with the vertical as shown in the figure.

For a complete circle, the minimum velocity at L must be \[v_L = \sqrt{5gl}\] .

Applying the law of conservation of energy, we have:

Total energy at M = total energy at L

Total energy at M = total energy at L

\[\text{ i .e } . , \frac{1}{2}m {v_M}^2 + mgl = \frac{1}{2}m {v_L}^2 \]

\[ \Rightarrow \frac{1}{2}m {v_M}^2 = \frac{1}{2}m {v_L}^2 - mgl\]

\[\text{ Using } v_L \geq \sqrt{5gl}, \text{ we have} : \]

\[\frac{1}{2}m {v_M}^2 \geq \frac{1}{2}m(5gl) - mgl\]

\[ \therefore v_M = \sqrt{3gl}\]

\[ \Rightarrow \frac{1}{2}m {v_M}^2 = \frac{1}{2}m {v_L}^2 - mgl\]

\[\text{ Using } v_L \geq \sqrt{5gl}, \text{ we have} : \]

\[\frac{1}{2}m {v_M}^2 \geq \frac{1}{2}m(5gl) - mgl\]

\[ \therefore v_M = \sqrt{3gl}\]

Is there an error in this question or solution?

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