The speed of the object if a string is wrapped tightly around a fixed reel that has a rotational inertia of 0.0352 kg · m² and a radius of 12.5 cm after it has fallen 1.25 m is 1.97 m /s
∑ Fy = m ay
mg - T = ma
m = 423 g = 0.423 kg
g = 9.8 m / s²
( 0.423 * 9.8 ) - T = 0.423 a
4.15 - T = 0.423 a → ( 1 )
τ = F * ⊥d
τ = I α
α = a / r
τ = Torque
F = Force
⊥d = Perpendicular distance
I = Moment of inertia
α = Angular acceleration
a = Tangential acceleration
r = Radius
I = 0.0352 kg m²
r = 12.5 cm = 0.125 m
F * ⊥d = I α
T * r = I a / r
T = I a / r²
T = 0.0352 a / 0.125²
T = 2.26 a → ( 2 )
( 1 ) becomes,
4.15 - 2.26 a = 0.423 a
2.683 a = 4.15
a = 1.55 m / s²
v² = u² + 2 a s
v = Final velocity
u = Initial velocity
s = Distance
u = 0
s = 1.25 m
v² = 0 + ( 2 * 1.55 * 1.25 )
v² = 3.875
v = 1.97 m / s
Therefore, the speed of the object after it has fallen 1.25 m is 1.97 m / s
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