# Aflywheel in the form of a heavy circular disk of diameter 0.612 m and mass 301 kg is mounted on a frictionless

###### Question:

## Answers

a) I= 56.37 kg m², b) W = 3.47 10⁵ J

c) ΔK = - 2.69 10⁵ J

Explanation:

a) The moment of inertia of the steering wheel is the moment of inertia of a disc with a shaft that passes through its center

I = ½ M R²

I = ½ 301 0.612²

I= 56.37 kg m²

b) Work is equal to the change in kinetic energy

W = ΔK = ½ I w² - ½ I w₀²

As part of the rest the initial angular velocity is zero

W = ½ I w²

Let's reduce the angular velocity to SI units

w = 1060 rev / min (2π rad / 1rev) (1min / 60s) = 111 rad / s

W = ½ 56.37 111²

W = 3.47 10⁵ J

c) Dissipated energy is equal to work

W = ΔK

W = ½ I [tex]w_{f}[/tex]² - ½ I w₀²

For this case

w₀ = 111 rad / s

[tex]w_{f}[/tex] = 502 rav / min (2pirad / 1rev) (1min / 60s) = 52,569 rad / s

ΔK = ½ 56.37 (52,569² - 111²)

ΔK = - 2.69 10⁵ J

B, if it was a lower watt she could put it in for her normal time and not burn it

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2