Question

What is the minimum velocity with which a body of mass m must enter a vertical loop of radius R so that it can complete the loop.
$\begin{align}
  & a)\sqrt{gR} \\
 & b)\sqrt{2gR} \\
 & c)\sqrt{3gR} \\
 & d)\sqrt{5gR} \\
\end{align}$

Answer 1

Hint: To make sure the body of mass m completes the entire loop we have to ensure that there is sufficient kinetic energy given to the body at the bottom so that it reaches the top of the circular loop. At the highest point, the tension on the body becomes zero and the gravitational force becomes equal to the centripetal force. Hence we will use the law of conservation of energy in order to determine the minimum velocity with which the body should enter the loop at the bottom.
Formula used:
${{F}_{CEP}}=\dfrac{m{{v}^{2}}}{R}$
$F=mg$
$K.E=\dfrac{1}{2}m{{v}^{2}}$

Complete step-by-step solution:

In the above figure, we can see a circle of radius R. Let us say we make sure the particle of mass ‘m’ enters with velocity ${{v}_{B}}$ at the bottom of the circle as shown below. Therefore the kinetic energy of the body while entering is equal to,
$K.E=\dfrac{1}{2}m{{v}_{B}}^{2}$
The above kinetic energy will be used to overcome the gravitational potential when the body reaches the top. When the body is at the top its gravitational force (F=mg) becomes equal to the centripetal force. Let us say the velocity of the body at the top is equal to ${{v}_{T}}$ and since centripetal force for a body moving in a circular path is given by ${{F}_{CEP}}=\dfrac{m{{v}^{2}}}{R}$ we can write,
$\begin{align}
  & {{F}_{CEP}}=F \\
 & \Rightarrow \dfrac{m{{v}_{T}}^{2}}{R}=mg \\
\end{align}$
The above body of mass m has to increase its height by 2R so in order to reach the top. Therefore the gravitational potential energy (U) which is expended for this purpose is given by U=mg2R.
Using law of conservation of energy for the above body we can write,
$\begin{align}
&\dfrac{1}{2}m{{v}_{B}}^{2}=mg2R+\dfrac{1}{2}m{{v}_{T}}^{2}\text{,}\because m{{v}_{T}}^{2}=Rmg \\
 & \Rightarrow \dfrac{1}{2}m{{v}_{B}}^{2}=mg2R+\dfrac{1}{2}Rmg \\
 & \Rightarrow \dfrac{1}{2}{{v}_{B}}^{2}=\dfrac{1}{2}\left[ 4gR+Rg \right] \\
 & \Rightarrow {{v}_{B}}^{2}=5gR \\
 & \Rightarrow {{v}_{B}}=\sqrt{5gR} \\
\end{align}$
Therefore the correct answer of the above question is option d.

Note: The basic idea of solving the above question is nothing but conservation of energy. It is to be noted that an above solution is a form of the answer. We can express it in various forms but depending on the given alternatives provided one has to make sure we can reduce it to a particular form.