Question

E
Two compound pendulums are made of:
(a) A disc of radius r and
(b) A uniform rod of length L. Find the minimum possible time period and distance between centre and
point of suspension in each case.​

Answer 1

Answer:

Part i)

For disc of radius "r" the distance of hinge from center is

$x = k = \frac{r}{\sqrt2}$

minimum time period is given as

$T = 2\pi \sqrt{\frac{\sqrt2 r}{g}}$

ii)

For rod of length L the distance of hinge from its center is

$k = \frac{L}{\sqrt{12}}$

minimum time period is given as

$T = 2\pi \sqrt{\frac{L}{\sqrt3 g}}$

Explanation:

As we know that time period of compound pendulum is given as

$\tau = I\alpha$

$\tau = mg x sin\theta$

here we know that by parallel axis theorem we have

$I = mk^2 + mx^2$

now we have

$mgx sin\theta = (mk^2 + mx^2) \alpha$

so we have

$\alpha = \frac{gx}{k^2 + x^2} \theta$

so we have

$T = 2\pi \sqrt{\frac{k^2/x + x}{g}}$

now for minimum value of time period we know that

$-\frac{k^2}{x^2} + 1 = 0$

$x = k$

so minimum time period is given as

$T = 2\pi \sqrt{\frac{2k}{g}}$

i) For disc we know that

$\frac{1}{2}mR^2 = mk^2$

$k = \frac{r}{\sqrt2}$

minimum time period is given as

$T = 2\pi \sqrt{\frac{\sqrt2 r}{g}}$

ii) For rod we have

$\frac{mL^2}{12} = mk^2$

$k = \frac{L}{\sqrt{12}}$

minimum time period is given as

$T = 2\pi \sqrt{\frac{L}{\sqrt3 g}}$

#Learn

Topic : Compound pendulum

https://brainly.in/question/7624929

Answer 2

a) 2πr2–√g−−−−√,r/22πr2g,r/2, b) 2πL3–√g−−−−√,L23–√2πL3g,L23.

Explanation:

(a) A disc of radius r and

2πr2–√g−−−−√,r/22πr2g,r/2

(b)  A uniform rod of length L

2πL3–√g−−−−√,L23–√2πL3g,L23.

To learn more:

1) Physical meaning of equivalent length of compound pendulum:

https://brainly.in/question/4464942

2) Difference between compound pendulum and torsional pendulum

https://brainly.in/question/1915528