91Ó°ÊÓ

given in the question,

wheel radius =0.280m

mass of suspended object =4.20kg

the substandard object moves downward a distance.

we know that \(y - {y_0} = \left( {\frac{{{v_{initial}} + {v_{final}}}}{2}} \right)t\) and e also know that,

\({K_{initial}} + {U_{initial}} = {K_{final}} + {U_{final}}\)

Here we find the angular velocity,

\(y - {y_0} = \left( {\frac{{{v_{initial}} + {v_{final}}}}{2}} \right)t\)

In the given question\(y = 3m\), \({y_0} = 0m\) and \({v_{{\mathop{\rm in}\nolimits} itial}} = 0\), \(t = 2\).

\(3m = \left( {{v_{final}}} \right)\left( {1s} \right)\)

We know the angular velocity of the rotation.

\({v_{linear}} = r\omega \)

\(3\frac{m}{8} = \left( {0.280m} \right)\omega \)

\(\frac{{3\frac{m}{s}}}{{0.28m}} = \omega \)

\(10.71\frac{{rad}}{s} = \omega \)

we know that \({K_{initial}} + {U_{initial}} = {K_{final}} + {U_{final}}\)

\(0 + {U_{initial}} = {K_{final}} + {U_{final}}\)

\(m.g.y = \frac{1}{2}I{\omega ^2} + \frac{1}{2}m{v^2}\)

\(m \times \left( {9.81\frac{m}{{{s^2}}}} \right)\left( {3m} \right) - \frac{1}{2}m{v^2} = \frac{1}{2}I{\omega ^2}\)

\(\frac{2}{{{\omega ^2}}}\left( {m \times \left( {9.81\frac{m}{{{s^2}}}} \right)\left( {3m} \right)} \right) = 1\)

\(\frac{2}{{{{\left( {10.71\frac{{rad}}{s}} \right)}^2}}}\left( {\left( {4.20kg} \right)\left( {9.81\frac{m}{{{s^2}}}} \right)\left( {3m} \right) - \frac{1}{2}\left( {9.81\frac{m}{{{s^2}}}} \right)} \right) = 1\)

\(1 \approx 1.82kg.{m^2}\)

We know that the formula ,

\(I = \frac{1}{2}m{r^2}\)

\(1.824kg.{m^2} = \frac{1}{2}m{r^2}\)

\(1.824kg.{m^2} = \frac{1}{2}m{\left( {0.280m} \right)^2}\)

\(m = \frac{{3.648kg.{m^2}}}{{0.078{m^2}}}\)

\(m = 46.5kg\).

Hence, the mass of the wheel is\(46.5kg\).