Question

A 20-cm-long, 190 g rod is pivoted at one end. A 19 g ball of clay is stuck on the other end.

What is the period if the rod and clay swing as a pendulum?

Answer 1

Length of the rod, L= 20cm=0.2m

Mass of the rod, M= 190g= 0.19kg

Mass of ball, m= 19g=0.019kg

Let Time period of the swing be "T".

Let $\omega$ be the angular frequency of the swing, $\theta$ be the angular frequency, then we can write the equation of motion of the rod about the pivot as,

$(\frac{ML}{2}+mL)gsin \theta= -I\frac{d^2\theta}{dt^2}$......................(1)

where g is the acceleration due to gravity, I be the total moment of inertia of the rod about the pivot and $\theta$ be the small angular displacement of the rod.  

Then we can write moment of inertia of the rod as,

$I= \frac{ML^2}{3}+mL^2$

Using above equation in equation 1,

  $(\frac{ML}{2}+mL)gsin \theta=-(\frac{ML^2}{3}+mL^2) \frac{d^2\theta}{dt^2}$

$(\frac{M}{2}+m)Lgsin \theta=-(\frac{M}{3}+m) L^2\frac{d^2\theta}{dt^2}$

Using small angle approximation: $sin \theta\approx \theta$ , we get

$(\frac{M+2m}{2})g\theta=-(\frac{M+3m}{3}) L\frac{d^2\theta}{dt^2}$

$\frac{d^2\theta}{dt^2}= -\frac{3(M+2m)g}{2L(M+3m)}\theta$

Now comparing above equation with standard equation of motion, $\frac{d^2\theta}{dt^2}= -\omega^2 \theta$

We get,

$\omega= \sqrt{\frac{3(M+2m)g}{2L(M+3m)}}$

Using given values in above equation,

$\omega= \sqrt{\frac{3[0.19+2(0.019)]9.8}{2\times 0.2[0.19+3(0.019)]}}$

$\omega= \sqrt{\frac{6.70}{0.0988}}= 8.23 rad/s$

Then Time Period of the swing will be

$T= \frac{ 2 \pi }{\omega}$

Using the value of angular frequency in above equation,

$T= \frac{ 2 \times 3.14 }{8.23}= 0.763s$(ANS)

$T= \frac{2.512 \times 0.247}{3\times 0.228 \times 9.8}=0.092s$(ANS)