Question

The figure here shows a smooth ‘looping-the-loop’ track. A particle, of mass m, is released from point A, as shown in the figure. If H = 3r, would the particle ‘loop the loop’?

Answer 1

Step1: Speed

1. The distance traveled by an object in a unit of time is known as speed.
2. Speed is a scalar quantity as it just considers the magnitude of the object and ignores its direction.

Step2: Given data

$H=3r$

Step3: Formula used

1. Loss in potential energy is determined as, $Energy=mgH[wherem=mass,H=height]$
2. The gain in kinetic energy is determined as, $Energy=\frac{1}{2}m{v}^2[wherem=mass,v=velocity]$

Step4: Calculating the speed

Let us assume that the speed acquired by the particle at the (lowest) point B is $v_B$.

According to the law of conservation of energy

$LossinpotentialenergyatA=GaininkineticenergyatB$

$$\begin{gathered} (0+mgH)=\frac{1}{2}m{v_B}^2+0 \\ mgH=\frac{1}{2}m{v_B}^2[ValueofH=3r] \\ mgetscancelledoutfromboththesides \\ 2g\left(3r\right)=m{v_B}^2 \\ 6gr={v_B}^2 \\ {v}_B=\sqrt{6gr} \end{gathered}$$

At point B the minimum speed is needed by the particle so that it can ‘loop the loop’ is $\sqrt{5gr}$.

Step2: Would the particle ‘loop-the-loop’

As we have calculated $v_{B>}\sqrt{5gr}$

Hence, the particle would ‘loop-the-loop’.