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A String is Wrapped on a Wheel of Moment of Inertia 0⋅20 Kg-m2 and Radius 10 Cm and Goes Through a Light Pulley to Support a Block of Mass 2⋅0 Kg as Shown in the Following Figure.

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प्रश्न

A string is wrapped on a wheel of moment of inertia 0⋅20 kg-m2 and radius 10 cm and goes through a light pulley to support a block of mass 2⋅0 kg as shown in the following figure. Find the acceleration of the block.

योग
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उत्तर

Moment of inertia of the bigger pulley, I = 0.20 kg-m2
r = 10 cm = 0.1 m,

Smaller pulley is light. Therefore, on neglecting its moment of inertia, we have

Mass of the block, m = 2 kg

From the free body diagram, we get

\[mg - T = ma........(1)\]

\[Tr = I\alpha\] And

\[a = \alpha r\]

\[\Rightarrow T = \frac{Ia}{r^2}  ........(2)\]

Using equations (1) and (2), we get

\[mg = \left( m + \frac{I}{r^2} \right)  a\]

\[ \Rightarrow   a = \frac{mg}{m + \frac{I}{r^2}}\]

\[= \frac{2 \times 9 . 8}{2 + \left( \frac{0 . 2}{0 . 01} \right)}\]

\[= \frac{19 . 6}{22} = 0 . 89  m/s^2\]

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अध्याय 10: Rotational Mechanics - Exercise [पृष्ठ १९७]

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एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
अध्याय 10 Rotational Mechanics
Exercise | Q 34 | पृष्ठ १९७

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