Question-175059

Question Number 175059 by mr W last updated on 17/Aug/22

Commented by mr W last updated on 17/Aug/22

a s o l i d h e m i s p h e r e w i t h m a s s M a n d

r a d i u s R i s c o n n e c t e d w i t h a t h i n r o d

w i t h m a s s m a n d l e n g t h L a t t h e

c e n t e r .

f i n d t h e p e r i o d o f s m a l l o s c i l l a t i o n s .

Commented by mr W last updated on 18/Aug/22

Answered by mr W last updated on 18/Aug/22

Commented by mr W last updated on 18/Aug/22

O A = \frac{L}{2}

O B = \frac{3 R}{8}

e = O C

e + \frac{L}{2} = \frac{M}{m + M} \times (\frac{L}{2} + \frac{3 R}{8})

e = \frac{M}{m + M} \times (\frac{L}{2} + \frac{3 R}{8}) - \frac{L}{2} > 0

I_{P} = \frac{{m L}^{2}}{12} + m {(R + \frac{L}{2})}^{2} + \frac{83 {M R}^{2}}{320} + M {(\frac{5 R}{8})}^{2}

I_{P} = m (\frac{L^{2}}{3} + R^{2} + R L) + \frac{13 {M R}^{2}}{20}

Commented by mr W last updated on 18/Aug/22

Commented by mr W last updated on 18/Aug/22

I_{p} {(\frac{d θ}{d t})}^{2} + (m + M) g (R - e \cos θ) = E = c o n s t a n t

2 I_{p} (\frac{d θ}{d t}) \frac{d^{2} θ}{{d t}^{2}} + (m + M) g e \sin θ \frac{d θ}{d t} = 0

2 I_{p} \frac{d^{2} θ}{{d t}^{2}} + (m + M) g e \sin θ = 0

f o r s m a l l θ, \sin θ \approx θ

\Rightarrow \frac{d^{2} θ}{{d t}^{2}} + \frac{(m + M) e g}{2 I_{P}} θ = 0 \to d . e . o f s . h . m .

ω = \sqrt{\frac{(m + M) e g}{2 I_{P}}} = \frac{1}{2} \sqrt{\frac{[4 (M - m) L + 3 M R] g}{2 m (\frac{L^{2}}{3} + R^{2} + R L) + \frac{13 {M R}^{2}}{10}}}

T = \frac{2 π}{ω} = 4 π \sqrt{\frac{20 m (\frac{L^{2}}{3} + R^{2} + R L) + 13 {M R}^{2}}{10 [4 (M - m) L + 3 M R] g}}

Commented by Tawa11 last updated on 20/Aug/22

Great sir

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