Answer :
The tangential speed of the outer edge of the cylinder at 5.82 s is approximately 10.571 m/s.
To find the tangential speed of the outer edge of the cylinder, we can use the equation:
v = r * ω
Where:
v is the tangential speed,
r is the radius of the cylinder, and
ω is the angular velocity.
Given:
Diameter of the cylinder = 35.9 cm = 0.359 m (since 1 m = 100 cm)
Radius of the cylinder, r = 0.359 m / 2 = 0.1795 m
Angular acceleration, α = 10.1 rad/s²
Time, t = 5.82 s
First, let's find the angular velocity at time t using the equation:
ω = ω₀ + α * t
Where:
ω₀ is the initial angular velocity (which is 0 since the cylinder is initially at rest).
ω = 0 + 10.1 rad/s² * 5.82 s
ω ≈ 58.842 rad/s
Now, we can substitute the values into the tangential speed equation:
v = r * ω
v = 0.1795 m * 58.842 rad/s
v ≈ 10.571 m/s
Therefore, the tangential speed of the outer edge of the cylinder at 5.82 s is approximately 10.571 m/s.
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