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

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

Therefore, a = 2ar

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

at = Rα equation 1

The radial acceleration is given as

ar = ω²R

Note that, at² + ar² = a²

at = √(a²-ar²)

Back to equation 1

at = Rα

Then, √(a²-ar²) = Rα

α = √(a²-ar²)/R

Where a = 2•ar

α = √((2•ar)²-ar²)/R

α = √(4ar²-ar²)/R

α = √3ar² / R

α = √3 • ar / R

Then, ar = ω²R

α = √3 • ω²R / R

α = √3 • ω²

answer: (431,2)

step-by-step explanation:

answer: a. same unit

explanation: