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.

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

poojatomarb76 poojatomarb76 · Answer 2 · 2022-04-15T07:58:24+02:00

The magnitude α of its angular acceleration at this instant if the acceleration vector a⃗ of a point on the rim of the wheel will be.[tex]\sqrt{3} \omega^2[/tex] rad/sec.

What is angular acceleration?

Angular acceleration is defined as the pace of change of angular velocity with reference to time.

The tangential velocity and the amount of the acceleration from a 30° angle. The tangential and the radial acceleration are found as;

[tex]\rm a_t = aCos30^0\\\\ \rm a_t = \sqrt{3} \frac{a}{2} \\\\ a_r = aSin30^0 \\\\ a_r =\frac{a}{2}\\\\ a=2a_r[/tex]

The relationship between the tangential acceleration and angular acceleration is found as;

[tex]a_t=R\alpha[/tex]

The radial acceleration in therm of the angular acceleration is found;

[tex]a_r= \omega^2 r[/tex]

On substituting the obtained values;

[tex]\sqrt{a^2-ar^2} =R \alpha \\\\ \alpha=\frac{\sqrt{a^2-ar^2}}{R} \\\\[/tex]

We got that

[tex]a=2a_r[/tex]

[tex]\sqrt{a^2-ar^2} =R \alpha \\\\ \alpha=\frac{\sqrt{(2a_r)^2-ar^2}}{R} \\\\ \alpha=\frac{\sqrt{(4a_r^2-ar^2}}{R} \\\\ \ \alpha=\frac{\sqrt{3}a^2R }{R} \\\\ \alpha=\sqrt{3} \frac{A_r}{R} \\\\ a_r = \omega^2R \\\\\ \alpha=\sqrt{3} \frac{\omega^2}{R} \\\\ \alpha=\sqrt{3} \omega^2R[/tex]

Hence the magnitude α of its angular acceleration will be [tex]\sqrt{3} \omega^2[/tex] rad/sec.

To learn more about angular acceleration refer to the link ;

https://brainly.com/question/408236