Answer:
The angular acceleration of the wheel is 3.357rad/s²
Explanation:
Given;
Diameter of the wheel, d = 0.8 m
radius of the wheel, r = 0.4 m
initial velocity of the wheel, u = 3 m/s
final velocity of the wheel, v = 6 m/s
number of revolutions of the wheel, N = 4 rev.
Convert the linear velocity in m/s to angular velocity in rad/s;
v = ωr
ω = v / r
initial angular velocity, [tex]\omega_i = \frac{3}{0.4} = 7.5 \ rad/s[/tex]
final angular velocity, [tex]\omega_f= \frac{6}{0.4} = 15\ rad/s[/tex]
The angular acceleration of the wheel is calculated as;
[tex]\omega _f^2 = \omega_i^2 + 2 \alpha \theta[/tex]
where;
α is the angular acceleration (rad / s²)
θ is the angular rotation (rad)
θ = Number of revolutions x 2π rad/rev
= 4 rev. x 2π rad/rev
= 25.136 rad.
[tex]\omega _f^2 = \omega_i^2 + 2 \alpha \theta\\\\2 \alpha \theta = \omega _f^2 - \omega_i^2\\\\\alpha = \frac{\omega _f^2 - \omega_i^2}{2 \theta} \\\\\alpha = \frac{15^2 -7.5^2}{2*25.136} \\\\\alpha = 3.357 \ rad/s^2[/tex]
α = 3.357 rad/s²
Therefore, the angular acceleration of the wheel is 3.357rad/s²
The angular acceleration of the wheel is 3.36 rad/s²
The diameter is 0.8 m, hence:
radius = diameter / 2 = 0.8/2 = 0.4 m
The initial angular speed (ω₀) = v / r = 3 m/s / 0.4 m = 7.5 rad/s
The final angular speed (ω) = v / r = 6 m/s / 0.4 m = 15 rad/s
The angular displacement (θ) = 4 rev * 2π = 8π
ω² = ω₀ + 2αθ
15² = 7.5² + 2(8π)α
α = 3.36 rad/s²
The angular acceleration of the wheel is 3.36 rad/s²
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