Answer:
KE=55.18 J
Explanation:
The angular velocity
ω [tex]=\frac{v}{r}=\frac{3.05\frac{m}{s}}{0.343m}[/tex]
The moment of inertia of one solid disk bicycle wheel is
[tex]I = \frac{1}{2}(M_w)r^{2}[/tex]
And the rotational kinetic energy of one wheel is
[tex]KE_w = \frac{1}{2}*I*w^2 \\KE_w = \frac{1}{2}*\frac{1}{2}*(M_w)*r^2*(\frac{v}{r})^2\\KE_w =\frac{1}{4}*(M_w)*v^2[/tex]
The total kinetic energy is then that of the frame and wheels plus the rotational kinetic energy.
[tex]KE = \frac{1}{2}*(M_f + 2*M_w)*v^2 + 2*\frac{1}{4} *(M_w)*v^2[/tex]
There is tow kinetic energy because are two wheels
Resolve
[tex]KE = \frac{1}{2} *(M_f + 3*M_w)*v^2[/tex]
[tex]KE = \frac{1}{2} *(6.55 kg + 3*0.820 kg)*(3.50 \frac{m}{s} )^2[/tex]
[tex]KE = 65.18 J[/tex]
