Question

The wheel has a weight of 5.50 lb, a radius of r=13.0 in, and is rolling in such a way that the center hub, O, is moving to the right at a constant speed of v=17.0 ft/s. Assume all the mass is evenly distributed at the outer radius r of the wheel/tire assembly. What is the total kinetic energy of the bicycle wheel?

Answer 1

`E_(k)=1589.5ftlb`

Step-by-step explanation:

`E_(k)=E_(movement)+E_(rotational)\\`

`E_(k)=(1)/(2)mv^(2)+(1)/(2)Iw^(2)` (1)

For this wheel:

`w=(v)/(r)`

`I=mr^(2)`: inertia of a ring

We replace (2) and (3) in (1):

`E_(k)=(1)/(2)mv^(2)+(1)/(2)(mr^(2))((v)/(r))^(2)=mv^(2)=5.5*17^(2)=1589.5ftlb`