Question

A small cube with mass $M$ starts at rest point $1$ at a height $4R$, where $R$ is the radius of the circular part of the track. The cube slides down the frictionless track and around the loop. The force that the track exerts on the cube at point $2$ is nearly __________ times the cube's weight $Mg$.

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is C $3$

Velocity at bottom most point$=\sqrt{8}gr$

Now$,$ at $2$ its displacement from bottom most point is $2R,$

The velocity at $2$ will be$:-$

$\begin{array}{c}{v}^2=u^2+2as\\ v^2=8gR+2\left(-g\right)\left(2R\right)\\ v^2=4gR\\ Now,netforceat2=F+mg=\frac{m{v}^2}{R}\\ F=\frac{m\left(4gR\right)}{R}-mgf=3mg\end{array}$

Hence,

option $\left(C\right)$ is correct answer.