Solution:
initial sphere mvr = final sphere mvr + Iω
where I = mL²/3 = 2.3g * (2m)² / 3 = 3.07 kg·m²
0.25kg * (12.5 + 9.5)m/s * (4/5)2m = 3.07 kg·m² * ω
where: ω = 2.87 rad/s
So for the rod, initial E = KE = ½Iω² = ½ * 3.07kg·m² * (2.87rad/s)²
E = 12.64 J becomes PE = mgh, so
12.64 J = 2.3 kg * 9.8m/s² * h
h = 0.29 m
h = L(1 - cosΘ) → where here L is the distance to the CM
0.03m = 1m(1 - cosΘ) = 1m - 1m*cosΘ
Θ = arccos((1-0.29)/1) = 44.77 º
Condensation.
iGreen, more like iNOTGreen amirite?
Answer:
Explanation:
The equation for this, since we are talking about weight on an elevator, is Newton's 2nd Law adjusted to fit our needs:
where the Normal Force needed to lift that elevator car is the tension. So the equation then becomes
T = ma + w where T is the tension in the cable to lift the elevator, m is the mass of the elevator (which we have to solve for), a is the acceleration of the elevator (positive since it's going up), and w is the weight of the elevator (which we have as 5500 N). Solving first for mass:
w = mg and
5500 =- m(10) so
m = 550 kg. Now we have what we need to solve for the tension:
T = 550(4.0) + 5500 and
T = 2200 + 5500 so
T = 7700 N