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for a point on the rim of the flywheel, what is the magnitude of the tangential acceleration after 2.00 s of acceleration?

1 Answer

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Step-by-step explanation:

We can use the formula for tangential acceleration:

a_t = r * α

where a_t is the tangential acceleration, r is the radius of the flywheel, and α is the angular acceleration.

To find the angular acceleration, we can use the formula:

θ = 1/2 * α * t^2

where θ is the angular displacement, t is the time, and α is the angular acceleration.

Rearranging this formula to solve for α, we get:

α = 2θ / t^2

Substituting in the given values, we get:

α = 2 * (2π) / (2.00 s)^2

Simplifying, we get:

α = 1.57 rad/s^2

Substituting this into the formula for tangential acceleration, along with the given radius of the flywheel, we get:

a_t = (0.274 m) * (1.57 rad/s^2)

Simplifying, we get:

a_t = 0.431 m/s^2

Therefore, the magnitude of the tangential acceleration of a point on the rim of the flywheel after 2.00 s of acceleration is approximately 0.431 m/s^2.

User Tyronen
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