Tangential acceleration of a point on the rim of the flywheel during this spin-up process is 0.2548 m/s².
Tangential acceleration is defined as the rate of change of tangential velocity of the matter in the circular path.
Given,
Radius of flywheel (r) = 1.96 cm = 0.0196m
Angular acceleration (α)= 13.0 rad/s²
The tangential acceleration formula is at=rα
where, α is the angular acceleration, and r is the radius of the circle.
using the formula; at=rα = (13.0 rad/s²) (0.0196m) = 0.2548 m/s².
The tangential acceleration is 0.2548 m/s².
Learn more about the Tangential acceleration with the help of the following link:
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The bicyclist accelerates with magnitude <em>a</em> such that
25.0 m = 1/2 <em>a</em> (4.90 s)²
Solve for <em>a</em> :
<em>a</em> = (25.0 m) / (1/2 (4.90 s)²) ≈ 2.08 m/s²
Then her final speed is <em>v</em> such that
<em>v</em> ² - 0² = 2<em>a</em> (25.0 m)
Solve for <em>v</em> :
<em>v</em> = √(2 (2.08 m/s²) / (25.0 m)) ≈ 10.2 m/s
Convert to mph. If you know that 1 m ≈ 3.28 ft, then
(10.2 m/s) • (3.28 ft/m) • (1/5280 mi/ft) • (3600 s/h) ≈ 22.8 mi/h
