A homogeneous ball is suspended from a thread with a length equal to the radius of the ball. How many times is the period of small oscillations of this pendulum greater than that of a mathematical pendulum suspended at the same distance from the centre of gravity?

Question

anonymous · Answer

What is the meaning of mathematical pendulum - a point mass attached to a massless string?

anonymous · Answer

Hold on! 
Mathematically , time period of pendulum is $\huge{T=2\pi \sqrt{\frac{I}{mgL}}}$

where I is the moment of inertia about the point of suspension and L is the distance b/w point of suspension and c.m. of the sphere.

And if the size of sphere attached at the end is very small (mathematically zero), then 

$I=mL^{2}$

This gives $\huge{T=2\pi \sqrt{\frac{L}{g}}}$

However this time the sphere cannot be assumed as a point mass. So ,I will not be mL^2
Instead ,we have to use parallel axis theorem and calculate the moment of inertia about the point of suspension.|dw:1373529716491:dw| On the left is our given pendulum.

$\huge{I=m(2R)^{2}+\frac{2}{5}mR^{2}=\frac{22}{5}mR^{2}}$

This is the value of I for the given pendulum. So its time period will be(as given by first formula) will be, $ \huge{T=2\pi \sqrt {\frac{11R}{5g}}}$

And the time period for mathematical pendulum will be as given by second formula

$\huge{T=2\pi \sqrt{\frac{2R}{g}}}$

souvik · Answer

so the time period of this pendulum/that of mathematical pendulum= $\sqrt{11/10}$~1.05
yeah i got it...thanks..:)