A gyroscope flywheel of radius 1.96 cm is accelerated from rest at 13.0 rad/s2 until its angular speed is 2270 rev/min. (a) What

Question

1 year ago

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is the tangential acceleration of a point on the rim of the flywheel during this spin-up process

1 answer:

nika2105 [10] · Answer 1 · 2024-02-09T00:48:03+03:00

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.

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