Question

A 15cm long 200g rod is provided at one end. A 20g ball of clay is stuck on the other and what is the period of the rod and clay ball swing as a pendulum

Answer 1

$\huge\underline{\underline{\rm Answer}}$

$\large\red{\boxed{\rm T=4.85sec}}$

$\huge\underline{\underline{\rm Solution}}$

$\large\underline{\underline{\sf Given:}}$

• Pivot Lenght (l) = 15cm

• Mass of pivot ($\rm{m_1}$) = 200g or 0.2kg

• Mass of ball ($\rm{m_2}$)=20g or 0.02kg

$\large\underline{\underline{\sf To\:Find:}}$

• Time period of the rod and clay ball swing as a pendulum (T) = ?

_________________________________________

Moment of Inertia of rod about pivot =$\rm{\frac{m_1l^2}{3}}$

Moment of Inertia of clay ball about pivot =$\rm{m_2l^2}$

$\underline{\underline{\tt Combine\: moment\:of\:Inertia \:of\:pivot:}}$

$\large\implies{\rm I=\left(\frac{m_1}{3}+m_2\right)l^2}$

$\large\implies{\rm \left(\frac{0.2}{3}+0.02\right)(0.15)^2}$

$\large\implies{\rm \frac{0.2+0.06}{3}}$

$\Large\implies{\sf I=0.00195kg.m^2}$

$\underline{\underline{\tt Combined\:mass(M)=m_1+m_2}}$

$\large\implies{\rm M=0.2+0.02 }$

$\Large\implies{\sf M=0.22g }$

$\large\underline{\underline{\tt Time\:Period(T):}}$

$\large{\boxed{\rm T=2π \sqrt{\frac{l}{Mgl}}}}$

$\large\implies{\rm T=2π\sqrt{\frac{0.00195}{0.22×9.8×0.15}}}$

$\large\implies{\rm T= 2π×\sqrt{0.006} }$

$\large\implies{\rm T=2×\frac{22}{7}×0.077 }$

$\large\implies{\rm T=\frac{33.88}{7}}$

$\large\implies{\rm T=4.85sec}$

Hence ,

Time period of the rod and clay ball swing as a pendulum is 4.84 sec

$\large\red{\boxed{\rm T=4.85sec}}$

Answer 2

Answer:

4.84 seconds

Explanation:

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