Question

A simple pendulum of length $L$ carries a bob of mass $m$. If the breaking strength of the string is $2mg$. The maximum angular amplitude from the vertical can be

Answer 1

Step 1: Given data

The length of the pendulum is $L$.

The mass of the bob is $m$.

The breaking strength of the string is $2mg$.

Step 2: Angular amplitude and breaking strength

1. The angular amplitude is the angular displacement of the pendulum with respect to the hanging point.
2. The breaking strength is the limit of mass. If we increase the mass of the pendulum greater than the breaking strength, then the string will break.

Step 3: Diagram

Step 4: Energy conservation in a simple pendulum motion

In simple pendulum motion, the sum of kinetic and potential energy is conserved, i.e, $mgL(1-\cos \theta )=\frac{1}{2}m{v}^2$, where $L(1-\cos \theta )$ is the height of the pendulum at the highest angular amplitude, and g is the acceleration due to gravity.

Step 5: Finding the velocity of the pendulum

We know, tension is maximum at the lowest point, so considering all the forces at point B,

$T=mg+\frac{m{v}^2}{L}$, mg is the weight of the pendulum, T is the tension and $\frac{m{v}^2}{L}$is a centripetal force on the pendulum. v is the velocity of the pendulum. Now,

$$\begin{gathered} \frac{m{v}^2}{L}=T-mg=\left(2mg-mg\right) \\ or{v}^2=gL \\ orv=\sqrt{gL}..............\left(1\right) \end{gathered}$$

Step 6: Finding the maximum angular amplitude

As we know, from the conservation of energy, $mgL(1-\cos \theta )=\frac{1}{2}m{v}^2$. So,

$$\begin{gathered} mgL(1-\cos \theta )=\frac{1}{2}m{v}^2=\frac{1}{2}m{\left(\sqrt{gL}\right)}^2\left(From\left(1\right),v=\sqrt{gL}\right) \\ orgL(1-\cos \theta )=\frac{1}{2}{\left(\sqrt{gL}\right)}^2=\frac{1}{2}\times gL \\ or(1-\cos \theta )=\frac{1}{2} \\ or\cos \theta =\frac{1}{2} \\ or\theta ={60}^o \end{gathered}$$

Therefore, the maximum angular amplitude from the vertical is ${60}^o$.