To solve this problem it is necessary to apply the equations related to the description of the tangential and angular movement.
The displacement where the speed and acceleration is related is given by the equation:
[tex]x = v_0t+\frac{1}{2}at^2[/tex]
Where
[tex]v_0 =[/tex] Initial velocity (0 because start from rest)
t = time
a = Acceleration
We have angular acceleration but not tangential acceleration. Tangential acceleration can be obtained through the relationship
[tex]a = r\alpha \rightarrow r = \frac{diameter}{2}[/tex]
[tex]a = \frac{0.08}{2}(3)[/tex]
[tex]a = 0.12m/s^2[/tex]
And we have also that the displacement is
[tex]x = 6m[/tex]
Now replacing,
[tex]x = v_0t+\frac{1}{2}at^2[/tex]
[tex]6 = 0*t+\frac{1}{2}(0.12)t^2[/tex]
[tex]6 = \frac{1}{2}(0.12)t^2[/tex]
[tex]t = 10s[/tex]
Therefore will take the cord to unwind around 10s
Answer:
t = 10 s
Explanation:
Data
D = 8.0 cm = 0.08 m : diameter of the wheel
R = D/2 = (0.08)/2 = 0.04 m : Radio of the wheel
Lc = 6.0 m : length of the cord
ω₀ = 0
α = 3.0 rad/s²
Problem development
Lw : Length of the circunference of the wheel
Lw = 2πR = 2π(0.04) m = 0.2513 m = 1 revolution
θ : angular displacement
1 revolution = 0.2513 m , Lc = 6.0 m
θ = 6.0 m * (1 rev/0.2513 m) = 23.87 rev = 23.87*2π= 150 rad
Kinematics of the wheel
θ = ω₀*t + (1/2)(α)(t)²
150 = 0 + (1/2)(3)(t)²
300 = (3) (t)²
(t)² = 100
[tex]t = \sqrt{100}[/tex]
t = 10 s
