Time period of SHM

What i am doing wrong in this problem? Here A is the COM and P is the bottom most point about which i calculated torque so as to prevent consideration of (friction for without slipping conditions). C is centre of cylinder shell. Using 1 and 2 i am getting answer to be 2pi√2R/g . I got the inspiration to do this problem using bottom most point as from this problem "20" in the link:: problems: https://www.slideshare.net/mobile/miranteogbonna/soal-latihan1mekanika , solutions: https://www.slideshare.net/mobile/miranteogbonna/jawaban-soallatihan1mekanika

#Mechanics

Easy Math Editor

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`bold` or `__bold__`	bold
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Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

@Steven Chase sir @Mark Hennings sir pls help

Aryan Yadav - 1 year, 8 months ago

I am also not getting what u did wrong , problem 20 does seem like it would work in this situation too, @Steven Chase sir what do u think?

Kudo Shinichi - 1 year, 8 months ago

In the first picture, what is the correct option? According to me, the answer should be option 1.

Karan Chatrath - 1 year, 8 months ago

It would be nice to know if I am right or wrong.

Karan Chatrath - 1 year, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$