91Ó°ÊÓ

Given spokes with a negligible moment of inertia : \[{I_{spokes}} = 0\]

Assume the rope is massless: \[{m_{rope}} = 0\]

Weight of the HC: \[{\omega _{HC}} = 40.0N\]

The radius of the HC: \[{R_{HC}} = 0.25m\]

\[\omega \] is the angular speed of the HC,

We know that the rim of the hollow cylinder is,

\[{v_2} = {\omega _2}{R_{HC}}\]

\[{\omega _2} = \frac{{{v_2}}}{{{R_{HC}}}}\]

\[{m_{HC}} = \frac{{{\omega _{HC}}}}{g}\]

By energy conservation

\[{U_1} + {K_1} + {\omega _{others}} = {U_2} + {K_2}\]

Here it has no change in gravitational potential energy y.

Since,\[{I_{spoken}} = 0 \to {m_{spokes}} = 0\]

So, \[{m_{rope}} = 0\] (negligible change )

So, \[{U_1} = 0\] and \[{U_2} = 0\].

Here we analyse the kinetic energy\[{K_1} = 0\], since the system is at rest.

The rotational kinetic energy of the HC is given by

\[{K_2} = \frac{1}{2}{I_{HC}}{\omega ^2}\]

\[{K_2} = \frac{1}{2}\left( {{M_{HC}}R_{Hc}^2} \right)\left( {\frac{{v_2^2}}{{{R_{HC}}}}} \right)\]

\[ = \frac{1}{2}{M_{HC}}v_2^2\]

\[ = \frac{1}{2}\left( {\frac{{{\omega _{HC}}}}{g}} \right)v_2^2\]

Put the value on the above expression,

\[ = \frac{1}{2}\left( {\frac{{40.00}}{{9.80}}} \right){\left( {6.00} \right)^2}\]

\[ = 73.471\]

Here applying the second newton’s law of the rope.

\[\sum {{F_y}} = {m_{rope}}{a_y}\]

\[F - P = \left( 0 \right){a_y}\]

\[ = 0\]

\[F = P\]

The magnitude of the force at each end of the rope is the same

\[{\omega _{rope}} = - Fd + Pd\]

\[ = - Pd + Pd\]

\[ = 0\]

Here F has the opposite direction to the displacement d.

The work done by the force P on the cylinder is

\[{\omega _{others}} = Pd = P\left( {5.00} \right)\]

\[ = 5.00P\]

The work done by P becomes the rational kinetic energy of the hollow cylinder.

\[{\omega _{others}} = {K_2}\]

\[5.00P = 73.47\]

\[P = 14.7N\]

Hence , the value of P is \[14.7N\].