91影视

\(\begin{array}{l}{\rm{Diameter}},\;D = 42\;{\rm{cm}} = 0.42 \times {10^{ - 2}}{\rm{m}}\\{\rm{Radius,}}\;r = 0.42 \times {10^{ - 2}}{\rm{m}}\\{\rm{Height,}}\;h = \;58\;{\rm{m}}\end{array}\)

The relationship between the linear velocity and the angular velocity is \(V = r\omega \).

Where, V is linear velocity and \(\omega \)is angular velocity.

Apply conservation of energy,

\(\begin{array}{c}{K_1} + {U_1} = {K_2} + {U_2}\\Mgh + 0 = 0 + \frac{1}{2}M{V^2} + \frac{1}{2}I{\omega ^2} \cdot \cdot \cdot \cdot \cdot \cdot \left( 1 \right)\end{array}\)

Where,

\(\begin{array}{c}M = {m_{rim}} + 6{m_{spokes}}\\I = {I_{rim}} + {I_{spokes}} = {m_{rim}}{r^2} + 6\left( {\frac{{{m_{spokes}}{r^2}}}{3}} \right)\end{array}\)

\({m_{rim}} = \lambda \left( {2\pi r} \right),\;{m_{spokes}} = \lambda r\)

Substitute these values in 鈥�M鈥� and 鈥�I鈥�,

\(\begin{array}{c}M = \lambda \left( {2\pi r} \right) + 6\left( {\lambda r} \right)\\ = 2\lambda r\left( {\pi + 3} \right)\end{array}\)

\(\begin{array}{c}I = \lambda \left( {2\pi r} \right){r^2} + 6\left( {\frac{{\lambda r\left( {{r^2}} \right)}}{3}} \right)\\ = 2\lambda \pi {r^3} + 2\lambda {r^3}\\ = 2\lambda {r^3}\left( {\pi + 1} \right)\end{array}\)

Substitute M and Ivalue in Equation (1), to find angular velocity \(\omega \),

\(\begin{array}{c}2r\lambda \left( {\pi + 3} \right)gh = \frac{1}{2} \times 2r\lambda \left( {\pi + 3} \right) \times {\left( {r\omega } \right)^2} + \frac{1}{2} \times 2\lambda {r^3}\left( {\pi + 1} \right) \times {\omega ^2}\\\omega = \sqrt {\frac{{\left( {\pi + 3} \right)gh}}{{{r^2}\left( {\pi + 2} \right)}}} \\ = \sqrt {\frac{{\left( {\pi + 3} \right) \times 9.8 \times 58}}{{{{\left( {0.21 \times {{10}^{ - 2}}} \right)}^2}\left( {\pi + 2} \right)}}} \\ = 12400\;{\rm{rad/s}}\end{array}\)

To find the linear velocity,

\(\begin{array}{c}V = r\omega \\ = 0.21 \times {10^{ - 2}} \times 12400\\ = 26\;{\rm{m/s}}\end{array}\)

Hence, the velocity is 26m/s.