Question

Lesson:- Rotational Dynamics
Class :- 12 (Maharastra board)

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Answer 1

Answer:

Refer attachment for your Answer.

• While solving problems related to rotational motion , we need to consider a force which acts on body due to its virtue of rotational i.e mω²r.

• Other thing to be noted is to balance forces in required direction of equilibrium.

• π² = g
• tanθ = tan37° = 3/4

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Answer 2

Aɴsᴡᴇʀ

☞ ғɪʀsᴛ ᴏғ ᴀʟʟ ʟᴇᴛ ᴄᴜʟᴄᴜʟᴀᴛᴇ ʀᴀᴅɪᴜs ɪɴ

☞ ᴡʜɪᴄʜ ʀɪᴅᴇʀ ʀᴇᴠᴏʟᴠᴇ =ʀ

ʀ=2.1ᴍ+[ᴄᴏs(90-∅)]×1.5

2.1+[1.5×0.6]=2.1ᴍ+0.9

☞3.0ᴍ

[•°•ʀ=3ᴍ]

ᴘᴏsɪᴛɪᴠᴇ ᴇǫᴜɪʟɪʙʀɪᴜᴍ ʟᴇᴛ ʙᴀʟᴀɴᴄᴇ ғᴏʀᴄᴇs

1)Iɴ ᴠᴇʀᴛɪᴄᴀʟ ᴅɪʀ^ɴ: T cos∅=mg----------(i)

2)In horizontal ^n:T sin∅ =mw²R------------(ii)

Divide both Fq^n (ii)÷(i)

Tan ∅=w²r/9 =(2πm)²R

=4π²n²R/9

$\longmapsto\therefore \rm \: tan \theta = \tt\frac{ 4\cancel{\pi ^{2} }n ^{2} \cancel{r} }{ \cancel{9}} \\ \\ \rm \longmapsto{n}^{2} = \frac{tan \theta}{4r} \\ \\{ \boxed{ \boxed{ \boxed{ \boxed{ \tt \: n = \sqrt{ \frac{tan}{4r} } }}}}} \\ \rm \implies where \: n = frequency \\ \\ \leadsto \rm \: n = \sqrt{ \frac{ \cancel{3}}{4} } \times \frac{1}{4} \times \frac{1}{ \cancel{8}} \\ \\ \therefore \frac{tan \theta = tan}{37 \degree = \frac{3}{4} } \\ \\ \leadsto \: n = \sqrt{ \frac{1}{6} } \\ \\ \boxed{ \therefore \: n = \frac{1}{4} \: ren |see } \\ \\ \\ { \boxed{ \boxed{ \boxed{ \boxed{ \boxed{ \mathfrak{answer = no \: \: of \: \: frequency = \frac{1}{4}rev |see }}}}}}}$

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