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At a certain instant, a rotating turbine wheel of radius RR has angular speed ωω (measured in rad/srad/s). What must be the magnitude αα of its angular acceleration (measured in rad/s2rad/s2) at this instant if the acceleration vector a⃗ a→ of a point on the rim of the wheel makes an angle of exactly 30∘∘ with the velocity vector v⃗ v→ of that point? Express your answer in terms of some or all of the variables RRR and ωωomega.

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Answer:

Explanation:

The magnitude of the acceleration makes an angle of 30° with the tangential velocity.

Resolving the acceleration to tangential and radial acceleration

at = aCos30 = √3a/2

ar = aSin30 = ½a

a = 2•ar

Then, the tangential acceleration is the linear acceleration, so the relationship between the tangential acceleration and angular acceleration is given as:

at = Rα

Then, α = at/R

since at = √3a/2

Then, α = √3 at/2R, equation 1

The radial acceleration is given as

ar = ω²R

Note that, at² + ar² = a²

at = √(a²-ar²)

Back to equation 1

α = √3 at/2R

α = √3√(a²-ar²)/2R

α = √3√(a²-(w²R)²)/2R

α = √3(a²-w⁴R²) / 2R

Also, a = 2•ar = 2w²R

Then,

α = √3((2w²R)²-w⁴R²) / 2R

α = √3(4w⁴R²-w⁴R²) / 2R

α = √3(3w⁴R²) / 2R

α = √9w⁴R² / 2R

α = 3w²R / 2R

α = 3w²/2

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