Answer:
Explanation:
90 rpm = 90 / 60 rps
= 1.5 rps
= 1.5 x 2π rad /s
angular velocity of flywheel
ω = 3π rad /s
Let I be the moment of inertia of flywheel
kinetic energy = (1/2) I ω²
(1/2) I ω² = 10⁷ J
I = 2 x 10⁷ / ω²
=2 x 10⁷ / (3π)²
= 2.2538 x 10⁵ kg m²
Let radius of wheel be R
I = 1/2 M R² , M is mass of flywheel
= 1/2 πR² x t x d x R² , t is thickness , d is density of wheel .
1/2 πR⁴ x t x d = 2.2538 x 10⁵
R⁴ = 2 x 2.2538 x 10⁵ / πt d
= 4.5076 x 10⁵ / 3.14 x .1 x 7800
= 184
R= 3.683 m .
diameter = 7.366 m .
b ) centripetal accn required
= ω² R
= 9π² x 3.683
= 326.816 m /s²
Answer:
The first difference is fairly straightforward: other countries prefer other sports. While some aspects of American sports culture are universal there are other things that are sports-specific, such as the 7th-inning stretch at a baseball game or counting the steps of a basketball player after he fouls out.
Explanation:
A uniform thin solid door has height 2.20 m, width .870 m, and mass 23.0 kg. Find its moment of inertia for rotation on its hinges. Is any piece of data unnecessary? So far, I don't understand how to calculate moments of inertia for things like this at all. I can do a system of particles, but when it comes to any ridgid objects, such as this door or rods or cylinders, I don't get it. So basically I have no idea where to even start with this.
so A
