Question

For a particle moving in circular motion is vertical plane the maximum speed is u and the minimum speed is v. If the maximum speed is double then the minimum speed becomes
A. $2v$
B. $\sqrt {5{u^2} + {v^2}}$
C. $4v$
D. $\sqrt {3{u^2} + {v^2}}$

Answer 1

Use the expressions for the minimum and maximum speeds of an object in vertical circular motion. Then use the condition for the new maximum speed and compare all these speeds to determine the new minimum speed if the initial maximum speed is doubled.

Formulae used:
The minimum speed of the object in vertical circular motion is
${v_{\min }} = \sqrt {Rg}$ …… (1)
Here, ${v_{\min }}$ is the minimum speed of the object in circular motion, $R$ is the radius of the circle and $g$ is the acceleration due to gravity.
The maximum speed of the object in vertical circular motion is
${v_{\max }} = \sqrt {5Rg}$ …… (2)
Here, ${v_{\max }}$ is the maximum speed of the object in circular motion, $R$ is the radius of the circle and $g$ is the acceleration due to gravity.

Complete step by step answer:
We have given that the maximum speed of the particle in the vertical circular motion is $u$ and the minimum speed of the same particle in the vertical circular motion in a vertical circle of the same radius is $v$. Let $R$ is the radius of the vertical circle.
Substitute $v$ for ${v_{\min }}$ in equation (1).
$v = \sqrt {Rg}$
Substitute $u$ for ${v_{\max }}$ in equation (1).
$u = \sqrt {5Rg}$
Take the ratio of the minimum speed to the maximum speed of the particle.
$\dfrac{u}{v} = \dfrac{{\sqrt {5Rg} }}{{\sqrt {Rg} }}$
$\Rightarrow \dfrac{u}{v} = \sqrt 5$

Let $v'$ and $u'$ be the new minimum and maximum speeds of the particle respectively. We have given that the maximum speed of the particle is doubled. Then the expression for new maximum speed becomes
$u' = 2\sqrt {5Rg}$
The ratios of the minimum speed to the maximum speed of the particle is in proportion with the ratio of the new maximum and minimum speed of the particle.
$\dfrac{u}{v} = \dfrac{{u'}}{{v'}}$
Substitute $\sqrt 5$ for $\dfrac{u}{v}$ and $2\sqrt {5Rg}$ for $u'$ in the above equation.
$\sqrt 5 = \dfrac{{2\sqrt {5Rg} }}{{v'}}$
$\Rightarrow v' = \dfrac{{2\sqrt {5Rg} }}{{\sqrt 5 }}$
$\Rightarrow v' = 2\sqrt {Rg}$
Substitute $v$ for $\sqrt {Rg}$ in the above equation.
$\therefore v' = 2v$
Therefore, the minimum speed becomes $2v$.

Hence, the correct option is A.

Note: The students should be careful while taking the minimum and maximum values of the speeds of the particle in vertical circular motion. If these values are not taken correctly, the final answer for the minimum speed of the particle will also be incorrect.