Answer:
v = 1.08 m/s
Explanation:
What is the linear speed of the 0.0500-kg sphere as its passes through its lowest point?
The decrease in PE is
d = 80.0cm * 1 / 1000m = 0.80m
h = 0.80 m /2 = 0.40 m
ΔPE = m*g*h
ΔPE = (0.0500 - 0.0200)kg * 9.8m/s² * 0.400 m
ΔPE = 0.1176 J
The moment of inertia of the assembly is
I = 1/12*m*L² + (m1 + m2)*(L/2)²
I = 1/12*0.390kg*(0.800m)² + 0.0700kg*(0.400m)²
I = 0.032 kg·m²
KE = ½Iω²
0.1176 J = ½ * 0.032kg·m² * ω²
ω = 2.71 rad/s
v = ωr = 2.71 rad/s * 0.400m
The linear velocity
v = 1.08 m/s
Answer:
g' = g/9 = 1.09 m/s²
Explanation:
The magnitude of free fall acceleration at the surface of earth is given by the following formula:
g = GM/R² ----- equation 1
where,
g = free fall acceleration
G = Universal Gravitational Constant
M = Mass of Earth
R = Distance between the center of earth and the object
So, in our case,
R = R + 2 R = 3 R
Therefore,
g' = GM/(3R)²
g' = (1/9) GM/R²
using equation 1:
g' = g/9
g' = (9.8 m/s)/9
<u>g' = 1.09 m/s²</u>
