Answer to Question #197194 in Physics for gibson

Question

A bicycle wheel of mass 2.5kg and diameter of 80cm rotates at 500rev/min. For this

wheel, find

(a) angular velocity.

(b) moment of inertia about the axis of rotation

(c) energy it acquires

Answer 1

a) By definition, the angular velocity is:

$$\omega = 2\pi f$$

where $f = 500rev/min \approx 8.33Hz$. Thus:

$$\omega = 2\pi\cdot 8.33 \approx 52.3 rad/s$$

b) Suppose the wheel is a cylindrical shell. Than its moment of inertia is:

$$I = mr^2 = \dfrac{md^2}{4}$$

where $m = 2.5kg$ is the mass of the wheel, and $d = 80cm = 0.8m$ is its diameter. Thus:

$$I = \dfrac{2.5\cdot 0.8^2}{4} =0.4kg\cdot m^2$$

c) The energy is:

$$E = \dfrac{I\omega^2}{2} = \dfrac{0.4\cdot 52.3^2}{2} \approx 547J$$

Answer. a) 52.3 rad/s, b) 0.4 kg*m^2, c) 547 J.