Question

A block of mass $m$slides down a frictionless track, then around the inside of a circular loop-the-loop of radius $R$. From what minimum height $h$ must the block start to make it around without falling off? Give your answer as a multiple of $R$.

Short answer

The block must start at a minimum height h to make it around the loop without falling off is $h=\frac{5}{2}R$.

Step-by-step solution

Energy :  

A quantitative property that can be transmitted from an object so that it can do work.

Explanation : 

A mass block slides down a frictionless track then looping around the inside of a circular loop of radius R.

Mass potential energy at height h is $mgh$.Object KE moving with velocity v is $\frac{1}{2}m{v}^2$. The centripetal force acting on mass moving with speed v around a circle of radius R = $\frac{m{v}^2}{R}$.

At the loop's highest point, the minimum velocity required such that the mass does not without fall off is,

$$\begin{gathered} \frac{m{v}^2}{R}=mg \\ {v}^2=gR \end{gathered}$$

The loop's energy at the bottom is KE which is,

$$\begin{gathered} {\left(KE\right)}_{bottom}={\left(KE\right)}_{top}+{\left(PE\right)}_{top} \\ {\left(KE\right)}_{bottom}=\frac{1}{2}m{v}^2+mg\left(2R\right) \\ {\left(KE\right)}_{bottom}=\frac{1}{2}m\left(gR\right)+mg\left(2R\right) \\ {\left(KE\right)}_{bottom}=\frac{5}{2}mgR \end{gathered}$$

Since, the energy is conserved, this ${\left(KE\right)}_{bottom}=PEofmass$at height h from which the mass falls down.

$$\begin{gathered} mgh={\left(KE\right)}_{bottom}=\frac{5}{2}mgR \\ h=\frac{5}{2}R \end{gathered}$$