Question

The two pendulums shown in the figure below, each consist of a uniform solid ball of mass M supported by a rigid massless rod, but the ball for pendulum A is very tiny while the ball for pendulum B is much larger having a non-negligible radius R = 1.15 m. The length L in the figure is L = 1.6 m. Find the ratio of the ratio TB/TA of the period of pendulum B to the period of pendulum A.

Answer 1

Answer:

P = 2 pi (I0 / M g L)^1/2 for object with inertia of I0

Note that if I0 = M L^2 you have P = 2 pi (L / g)^1/2 the simple pendulum

Radius of sphere = 1.15 = .719 L

I0 = M L^2 + 2/5 M (.719 L)^2 = 1.207 M L^2

P2 = 2 pi (1.207 M L^2 / M g L)^1/2 = 2 pi (1.207 L / g)^1/2

P2 / P1 = (1.207 / 1)^1/2 = 1.10     ratio of periods

Answer 2

Answer:

the ratio is 46:101.

Explanation:

for pendulum 'B', periodic time;

$T_{B} =2\pi \sqrt{\frac{I}{MgL} }$

inertia of solid sphere$I=\frac{2}{5}MR^{2}$

I is Moment of Inertia, L is length of rod, and

periodic time for pendulum 'A';

$T_{A}=2\pi \sqrt{\frac{L}{g} }$

Thus;

$\frac{T_{B} }{T_{A} }=\frac{R}{L}\sqrt{\frac{2}{5} }$

$\frac{T_{B} }{T_{A} }=\frac{1.15}{1.6}\sqrt{\frac{2}{5} } =\frac{46}{101}$

What is periodic time?

The time taken in one complete oscillation is called time period.

What is moment of inertia?

The product of mass of particle(m) and square of the distance from rotational axis (r) is called moment of inertia.

$I=mr^{2}$

hence, the ratio is 46:101.

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