Question

A particle of mass m is tied to a string of length l on a smooth incline plane of inclination $\theta ={30}^{\circ }.$ Particle is imparted a velocity of $V_o$ at the bottom most point so that particle moves in a circle of radius l.

• A. Minimum value of $V_o$ so that particle complete circle is $\sqrt{\frac{5gl}{2}}$
• B. For $V_0=\sqrt{\frac{5gl}{2}}$ tension in string will never become zero
• C. Maximum tension in string if $V_o=\sqrt{\frac{5gl}{2}}$ is 3mg
• D. Minimum tension in string for $V_o=\sqrt{2gl}$ is not zero

Answer 1

Answer: (A), (C)

The correct options are
A Minimum value of $V_o$ so that particle complete circle is $\sqrt{\frac{5gl}{2}}$
C Maximum tension in string if $V_o=\sqrt{\frac{5gl}{2}}$ is 3mg


Take $g^1=gsin\theta =\frac{g}{2}$
At top posion,

T = 0 $\Rightarrow$
$\mathrm{m/}{g}^1=\frac{\mathrm{m/}{v}^2}{l}$
$v=\sqrt{g^{\prime }l}=\sqrt{\frac{g}{2}l}$
By conservation of energy, $\Rightarrow \frac{1}{2}\mathrm{m/}{v}^2+2\mathrm{m/}{g}^{\prime }l=\frac{1}{2}\mathrm{m/}{V}_o^2$
$\frac{g^{1}l}{2}+2g^{1}l=\frac{V_0^2}{2}\Rightarrow 5g^{1}l=V_0^2$
$V_0=\sqrt{5g^{1}l}=\sqrt{\frac{5}{2}gl}$
If the speed is $\sqrt{\frac{5}{2}gl}$, or below, the tension becomes zero.
The maximum tension occurs at the bottom most position.
$T=mg+\frac{m{V}_0^2}{l}$, when $V_0=\sqrt{\frac{5}{2}gl}$, T becomes $3mg$.