(a) Without the wheels, a bicycle frame has a mass of 6.75 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 3.95 m/s. J(b) Before the invention of a wheel turning on an axle, ancient people moved heavy loads by placing rollers under them. (Modern people use rollers, too. Any hardware store will sell you a roller bearing for a lazy susan.) A stone block of mass 675 kg moves forward at 0.395 m/s, supported by two uniform cylindrical tree trunks, each of mass 82.0 kg and radius 0.343 m. No slipping occurs between the block and the rollers or between the rollers and the ground. Find the total kinetic energy of the moving objects.

a) The total kinetic energy is that of the total mass of the bicycle plus the rotational kinetic energy of the two wheels. The linear speed of the circumference of the wheels matches the forward speed of the bicycle, so their angular speed is

ω = v/r

The moment of inertia of one solid disk bicycle wheel is

I = 0.5*m₂*r²

And the rotational kinetic energy of one wheel is

Kr = 0.5*I*ω² = 0.5*(0.5*m₂*r²)*(v/r)² = 0.25*m₂*v²

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

Ktotal = 0.5*(m₁ + 2*m₂)*v² + 2*(0.25*m₂*v²)

⇒ Ktotal = 0.5*v²*(m₁ + 3*m₂)

where

m₁ = 6.75 Kg

m₂ = 0.820 kg

v = 3.95 m/s

then

⇒ Ktotal = 0.5*(3.95 m/s)²*(6.75 Kg + 3*0.820 kg)

⇒ Ktotal = 71.85 J

b) We can apply the same equation obtained before

⇒ Ktotal = 0.5*v²*(m₁ + 3*m₂)

where

m₁ = 675 Kg

m₂ = 82.0 kg

v = 0.395 m/s

then

⇒ Ktotal = 0.5*(0.395 m/s)²*(675 Kg + 3*82 kg)

⇒ Ktotal = 71.85 J