Question

As shown in figure , a Bob of mass m is tied by massless string whose other end proportion is wound on a fly wheel (disc) of radius R and mass m when released from rest , the bob starts falling vertically . When it has covered distance h .The angular speed of wheel be : ​

JEE Mains 2020

Answer 1

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Answer 2

Answer:

• $\large{\bold{\underline{\omega = \dfrac{1}{R}\sqrt{\dfrac{4gh}{3}}}}}$ rad/sec

Given:

1. Radius = R.
2. Mass of Bob = M.
3. Distance covered by bob = h.

Explanation:

$\rule{300}{1.5}$

According to Law of Conservation of energy,

Work Done by the all the Force = Change in Rotational Kinetic energy.

$\large{\boxed{\bold{W_{(gravity)} = \Delta K.E_{(Rotational)}}}}$

$\bold{Here}\begin{cases}W_{\sf(gravity)} \text{represents Work Done by Gravity} \\ \\ \\ K.E_{\sf(Rotational)} \text{represents Rotational Kinetic energy}\end{cases}$

Substituting the values,

$\large{\tt \hookrightarrow mgh = \dfrac{1}{2} m v^2 + \dfrac{1}{2}I\omega^2}$

As we Know v = rω.

Substituting,

$\large{\tt \hookrightarrow mgh = \dfrac{1}{2} m (R \omega)^2 + \dfrac{1}{2}I\omega^2}$

$\large{\tt \hookrightarrow mgh = \dfrac{1}{2} m R^2 \omega^2 + \dfrac{1}{2}I\omega^2}$

Now,

I = MK²

• Where K is radius of Gyration.

$\large{\tt \hookrightarrow mgh = \dfrac{1}{2} m R^2 \omega^2 + \dfrac{1}{2} m K^2\omega^2}$

$\large{\tt \hookrightarrow mgh = \dfrac{1}{2} m \omega^2 \bigg[R^2 + K^2\bigg]}$

$\large{\tt \hookrightarrow \cancel{m}gh = \dfrac{1}{2} \cancel{m} \omega^2 \bigg[R^2 + K^2\bigg]}$

$\large{\tt \hookrightarrow gh = \dfrac{1}{2} \omega^2 \bigg[R^2 + K^2\bigg]}$

Now,

$\large{\tt \hookrightarrow gh = \dfrac{1}{2} \omega^2 R^2 \bigg[1 + \dfrac{K^2}{R^2}\bigg]}$

Now, we Know,

Moment of Inertia of Disc = MR²/2.

I = Mr²/2

• I = MK²

MK² = MR²/2

K² = R²/2

K²/R² = 1/2.

Substituting,

$\large{\tt \hookrightarrow gh = \dfrac{1}{2} \omega^2 R^2 \bigg[1 + \dfrac{1}{2}\bigg]}$

$\large{\tt \hookrightarrow gh = \dfrac{1}{2} \omega^2 R^2 \bigg[ \dfrac{2 + 1}{2}\bigg]}$

$\large{\tt \hookrightarrow gh = \dfrac{1}{2} \omega^2 R^2 \bigg[ \dfrac{3}{2}\bigg]}$

$\large{\tt \hookrightarrow gh = \dfrac{3}{4} R^2 \omega^2}$

$\large{\tt \hookrightarrow \dfrac{4gh}{3R^2} = \omega^2}$

$\large{\tt \hookrightarrow \omega^2 = \dfrac{4gh}{3R^2}}$

$\large{\tt \hookrightarrow \omega = \sqrt{\dfrac{4gh}{3R^2}}}$

$\large{\boxed{\boxed{\tt \omega = \dfrac{1}{r}\sqrt{\dfrac{4gh}{3}}}}}$

∴ Hence Derived!

$\rule{300}{1.5}$