91Ó°ÊÓ

Given in the question,

The Mass of the student,\({m_s} = 43.0\,\,{\rm{kg}}\)

The Height above the building\(h = \,12\,\,{\rm{m}}\)

The radius of the wheel\({r_w} = 0.300\,\,{\rm{m}}\)

The moment of inertia of the wheel, \(I = 9.60\,\,{\rm{kg}}{\rm{.}}{{\rm{m}}^{\rm{2}}}\)

Law of conservation of energy

According to the conservation of energy, the energy of a systemcan neither be created nor destroyed it can only be converted from one form of energy to another.

\({E_F} = {E_I}\)

From the conservation of energy,

Total initial energy = Total final energy

Initial potential energy +initial kinetic energy= final potential energy + final kinetic energy.

\({U_i} + K{E_i} = {U_f} + K{E_f}\)

Since the student is at height h.

The initial potential energy,\({U_i} = {m_s}gh\)

The system (student + wheel) is initially at the rest

The initial kinetic energy \(K{E_i} = 0\)

The final height is zero

Therefore,

Final potential energy \({U_f} = 0\)

The final kinetic energy will be the sum of the translational kinetic energy of the student and the rotational kinetic energy of the wheel.

Final kinetic energy

\(\begin{array}{}K{E_f} = K{E_{ts}} + K{E_{rw}}\\ = \frac{1}{2}{m_s}{v^2} + \frac{1}{2}I{\omega _w}^2\end{array}\)

Where v is linear speed, I is the moment of inertia and \(\omega \) is angular velocity.

Substituting all the values into the equation

\({m_s}gh + 0 = 0 + \frac{1}{2}{m_s}{v^2} + \frac{1}{2}{I_w}{\omega _w}^2\)

We know,

\(v = r\omega \)

\({m_s}gh = \frac{1}{2}{m_s}{\left( {{r_w}{\omega _w}} \right)^2} + \frac{1}{2}{I_w}{\omega _w}^2\)

Substituting the values

\(\begin{array}{}\left( {43.0\,\,{\rm{kg}}} \right)\left( {9.0\,\,{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)\left( {\,12\,\,{\rm{m}}} \right) = \frac{1}{2}\left( {43.0\,\,{\rm{kg}}} \right){\left( {0.300\,\,{\rm{m}}} \right)^2}{\omega _w}^2 + \left( {9.60\,\,{\rm{kg}}{\rm{.}}{{\rm{m}}^{\rm{2}}}} \right){\omega _w}^2\\{\omega _w}^2 = 750.70\,\,\\{\omega _w} = 27.4\,\,{{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}\end{array}\)

Therefore, the final angular velocity of the wheel is \({\omega _w} = 27.4\,\,{{{\rm{rad}}} \mathord{\left/ {\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}\)

Therefore, the final angular displacement

\(\begin{array}{}\theta = \frac{h}{r}\\ = \frac{{12.00\,\,{\rm{m}}\,}}{{0.300\,{\rm{m}}}}\\ = 40\,\,{\rm{rad}}\end{array}\)

To find the time first we need to find the angular acceleration since, the force acting on the system is constant so we can use the equation

\(\omega = \omega _0^2 + 2\alpha \left( {\theta - {\theta _0}} \right)\)

Where, \(\omega \) is final angular velocity =\({\omega _w} = 27.4\,\,{{{\rm{rad}}} \mathord{\left/ {\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}\)

Since the system is initially at rest initial angular velocity \({\omega _0} = 0\)

Initial angular displacement \({\theta _0} = 0\)

Final angular displacement \(\theta = 40\,\,{\rm{rad}}\)

Substituting into the equation.

\(\begin{array}{}{\left( {27.4\,\,{{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}} \right)^2} = 0 + 2\alpha \left( {40\,\,{\rm{rad}} - 0} \right)\\\alpha = 9.39\,\,{{{{\rm{rad}}} \mathord{\left/ {\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.}{\rm{s}}}^2}\end{array}\)

Therefore, angular acceleration is

Now for finding the time using the equation

\(\omega = {\omega _0} + \alpha t\)

Substituting the values

\(\begin{array}{}27.4\,\,{{{\rm{rad}}} \mathord{\left/ {\vphantom {{{\rm{rad}}} {\rm{s}}}} \right. } {\rm{s}}} = 0 + \left( {9.39\,\,{{{{{\rm{rad}}} \mathord{\left/ {\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}}^2}} \right)t\\t = 2.92\,\,{\rm{s}}\end{array}\)

Therefore, the time to reach the sidewalk is \(2.92\,\,{\rm{s}}\)

Since both the wheel and the student is connecting, they will be same transitional speed

So, the speed of a student can be calculated using the equation

\(\begin{array}{}v = r{\omega _w}\\ = \left( {0.300\,\,{\rm{m}}} \right)\left( {27.4\,\,{{{\rm{rad}}} \mathord{\left/ {\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}} \right)\\ = 8.22\,\,{{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}^2}\end{array}\)

Hence, she is moving at the speed of \( = 8.22\,\,{{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}^2}\)