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6.System of Particles and Rotational Motion
hard
$m$ દળનો પદાર્થ મુકત કરતાં $h$ અંતર કાપ્યા પછી તેનો વેગ
A
$\sqrt {2gh} $
B
$\sqrt {2gh} \frac{M}{m}$
C
$\sqrt {2gh\,m/M} $
D
$\sqrt {4mgh/2m + M} $
Solution
Let the acceleration of the mass be $a$.
Let the angular acceleration of the disc be $\alpha .$
since the string moves without slipping on the disc,
$a=R \alpha$
$a=\frac{m g-T}{m}$
$T=m g-m a…..(1)$
$\alpha=\frac{\tau}{I}$
where $\tau=$ torque $=T R$
and $I=\frac{M R^{2}}{2}$
$T=\frac{M a}{2}$
From $(1)$ and $(2)$
$a=\frac{2 m g}{m+2 M}$
${V}^{2}={u}^{2}+2as$
$u=0$
$s=h$
$V=\sqrt {\dfrac {4mgh}{2m+M}}$
Standard 11
Physics
