Question

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

(a) √(gl)
(b) √(2gl)
(c) √(3gl)
(d) √(5gl)

Answer 1

Answer ⇒ Option (c). √(3gl)

Solutions ⇒ 
Let the Speed of the Particles when the strings is vertical  be v m/s. 
Now, It is balanced by its own weight, i.e., mg.  
Therefore, mv²/l = mg 
∴ v²/l = g
⇒ v² = gl

 Let the speed of the Particles when the strings is horizontal be u m/s.

Now, 
Total Energy = (1/2)mu²
Also, Total Energy = (1/2)mv² + mgl

Thus, From this we get,
 (1/2)mu² = (1/2)mv² + mgl
(1/2)m(u² - v²) = mgl
⇒ u² - v² = 2gl.
∵ v² = gl,
∴ u² - gl = 2gl
⇒ u² = 2gl + gl
⇒ u² = 3gl
∴ u = √(3gl)


Hence, Option (c). is correct.


Hope it helps.