Question-108588

Question Number 108588 by ajfour last updated on 17/Aug/20

Commented by ajfour last updated on 21/Aug/20

T h e r e i s n o f r i c t i o n b e t w e e n b l o c k

a n d s p h e r e, b u t e n o u g h f r i c t i o n

b e t w e e n g r o u n d a n d s p h e r e, s u c h

t h a t s p h e r e r o l l s a s b l o c k s l i d e s

b a c k w a r d s . F i n d m a x i m u m

v e l o c i t y a c q u i r e d b y s p h e r e (h o l l o w) .

Answered by ajfour last updated on 21/Aug/20

Commented by ajfour last updated on 21/Aug/20

(N \sin θ) R (1 + \cos θ) = I α \dots (i)

A = α R

m g \cos θ - N - m α R \sin θ = \frac{{m v}^{2}}{R} . . (i i)

m g \sin θ + m α R \cos θ = \frac{m v d v}{R d θ} \dots (i i i)

\frac{v}{R} = \frac{d θ}{d t} \dots (i v)

\Rightarrow

m g \sin θ - \frac{I α}{R \sin θ (1 + \cos θ)}

- m α R \sin θ = m R {(\frac{d θ}{d t})}^{2}

f r o m (i i i)

R α = \frac{1}{\cos θ} (\frac{v d v}{R d θ} - g \sin θ)

a n d l e t \frac{I}{{m R}^{2}} = k, t h e n

g \sin θ - (\frac{k}{\sin θ (1 + \cos θ)} + \sin θ) α R

= R {(\frac{d θ}{d t})}^{2} \Rightarrow

\frac{d^{2} θ}{{d t}^{2}} - (\frac{g}{R}) \sin θ = \frac{[(\frac{g}{R}) \sin θ - {(\frac{d θ}{d t})}^{2}] \cos θ}{\sin θ + \frac{k}{\sin θ (1 + \cos θ)}}

\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots

Answered by mr W last updated on 22/Aug/20

Facebook Tweet Pin