The chain of length L is released from rest on a smooth fixed incline with x = 0 as shown in the figure. Determine velocity $v$ of the chain when a half of the length has fallen (Take $L = \frac{4}{25} m)$ in m/s. (Neglect edge effect of inclined). (Answer upto two digit after decimal places)

Open in App

Solution

We know that total energy is conserved
$ε_{i} = ε_{f} ε_{i} = i n i t i a l p o t e n t i a l e n e r g y + i n i t i a l k i n e t i c e n e r g y ε_{f} = f i n a l p o t e n t i a l e n e r g y + f i n a l k i n e t i + e n e r g y m g \frac{L}{2} s i n θ + 0 = m_{1} g \frac{(L - x)}{2} s i n θ - m_{2} g \frac{x}{2} + \frac{1}{2} m v^{2} N o w i f t h e m a s s i s c o n s i d e r e d a s u n i f o r m l y d i s t r i b u t e d t h e n m_{1} = \frac{m}{L} (L - x) m_{2} = \frac{m}{L} x m g \frac{L}{2} s i n θ + 0 = \frac{m}{L} (L - x) g \frac{(L - x)}{2} s i n θ - \frac{m}{L} x g \frac{x}{2} + \frac{1}{2} m v^{2} g L s i n θ = \frac{g}{L} (L^{2} + x^{2} - 2 L x) s i n θ - g \frac{x^{2}}{L} + v^{2} A s i t i s g i v e n h a l f o f t h e l e n g t h i s l e f t s o x = \frac{L}{2} v = \sqrt{\frac{5}{8} g l}$
Taking $g = 10 m / s^{2}$
Putting value of $L = \frac{4}{25}$

$v = 1 m / s$

Suggest Corrections

Similar questions

L

x = 0

v

L = \frac{4}{25} m)

m / s

v

L = \frac{4}{25} m)

l

m

l

m / s^{2}

m

Join BYJU'S Learning Program