Answer:
v = 6.195 m / s
Explanation:
For this exercise we can use the conservation of energy, for the system formed by the ball and the pulley
starting point. Higher before releasing the system
Em₀ = U = M g h
final point. When the ball has lowered h = 2
Em_f = K = ½ M v² + ½ I w²
the energy is preserved
Em₀ = Em_f
M g h = ½ M v² + ½ I w²
angular and linear velocity are related
v = w r
w = v / r
indicate that the moment of inertia is
I = ½ m r²
we substitute
M g h = ½ M v² + ½ (½ m r²) (v/r) ²
½ v² (M + [tex]\frac{1}{2}[/tex] m) = M g h
v² = [tex]2gh \ \frac{M}{M + \frac{m}{2} }[/tex]
let's calculate
v = [tex]\sqrt{ 2 \ 9.8 \ 2 \ \frac{70}{70 + 1.5} }[/tex]
v = 6.195 m / s
