Question

Without the wheels, a bicycle frame has a mass of 6.55 kg. Each of the wheels can be roughly modeled as a uniform solid disk with a mass of 0.820 kg and a radius of 0.343 m. Find the kinetic energy of the whole bicycle when it is moving forward at 3.05 m/s.

Answer 1

KE=55.18 J

Step-by-step explanation:

The angular velocity

ω
`=(v)/(r)=(3.05(m)/(s))/(0.343m)`

The moment of inertia of one solid disk bicycle wheel is

`I = (1)/(2)(M_w)r^(2)`

And the rotational kinetic energy of one wheel is

`KE_w = (1)/(2)*I*w^2 \\KE_w = (1)/(2)*(1)/(2)*(M_w)*r^2*((v)/(r))^2\\KE_w =(1)/(4)*(M_w)*v^2`

The total kinetic energy is then that of the frame and wheels plus the rotational kinetic energy.

`KE = (1)/(2)*(M_f + 2*M_w)*v^2 + 2*(1)/(4) *(M_w)*v^2`

There is tow kinetic energy because are two wheels

Resolve

`KE = (1)/(2) *(M_f + 3*M_w)*v^2`

`KE = (1)/(2) *(6.55 kg + 3*0.820 kg)*(3.50 (m)/(s) )^2`

`KE = 65.18 J`