Final answer:
The magnitude of the tangential acceleration of a bug on the rim of a 12-inch diameter disk, accelerating uniformly from rest to 80.0 rev/min in 4.10 seconds, is approximately 0.311 m/s².
Step-by-step explanation:
To calculate the magnitude of the tangential acceleration of a bug on the rim of a 12-inch diameter disk that accelerates uniformly from rest to an angular speed of 80.0 revolutions per minute (rev/min) in 4.10 seconds, we need to use the formulas of rotational motion. First, we convert the given angular speed to radians per second (rad/s) knowing that 1 rev/min is equal to π/30 rad/s. Then, we calculate the angular acceleration (α) using the formula α = ω/t, where ω is the angular speed and t is the time. Finally, we find the tangential acceleration (at) using the formula at = α×r, where r is the radius of the disk.
Conversion from rev/min to rad/s:
80.0 rev/min × (π rad/30 s/min) = 8π/3 rad/s.
Calculation of angular acceleration:
α = ω/t = (8π/3 rad/s) / 4.10 s ≈ 2.04 rad/s².
Calculation of tangential acceleration:
First, we convert the diameter to radius in meters: 12.0 inches / 2 = 6.0 inches ≈ 0.1524 m.
Then, at = α×r = 2.04 rad/s² × 0.1524 m ≈ 0.311 m/s².
The magnitude of the tangential acceleration of the bug is approximately 0.311 meters per second squared (m/s²).