Question

A 13-kg (including the mass of the wheels) bicycle has 1-m-diameter wheels, each with a mass of 3.1 kg. The mass of the rider is 38 kg. Estimate the fraction of the total kinetic energy of the rider-bicycle system is associated with rotation of the wheels?

Answer 1

`fraction = 0.11`

Step-by-step explanation:

Linear kinetic energy of the bicycle is given as

`KE = (1)/(2)mv^2`

`K_1 = (1)/(2)(13) v^2`

`K_1 = 6.5 v^2`

Now rotational kinetic energy of the wheels

`K_2 = 2((1)/(2)(I)(\omega^2))`

`K_2 = (mR^2)((v^2)/(R^2))`

`K_2 = mv^2`

`K_2 = 3.1 v^2`

now kinetic energy of the rider is given as

`K_3 = (1)/(2)Mv^2`

`K_3 = (1)/(2)(38) v^2`

`K_3 = 19 v^2`

So we have

`fraction = (K_2)/(K_1 + K_2 + K_3)`

`fraction = (3.1 v^2)/(6.5 v^2 + 3.1 v^2 + 19 v^2)`

`fraction = 0.11`