Answer to Question #174054 in Physics for Andu

Question #174054

Wheel with radius R = 10 cm, rotates with a constant angular acceleration = 3.14 rad / s². Find: a) angular velocity b) linear velocity c) tangential acceleration d) centripetal acceleration e) full acceleration f) the angle formed by the direction of acceleration of the point with the radius of the wheel. of the first second after the start of the movement.


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Expert's answer
2021-03-23T11:12:26-0400

(a) We can find the angular velocity as follows:

ω=ω0+αt,\omega=\omega_0+\alpha t,ω=0+3.14 rads21 s=3.14 rads.\omega=0+3.14\ \dfrac{rad}{s^2}\cdot1\ s=3.14\ \dfrac{rad}{s}.

(b) We can find the linear velocity as follows:


v=ωr=3.14 rads0.1 m=0.314 ms.v=\omega r=3.14\ \dfrac{rad}{s}\cdot0.1\ m=0.314\ \dfrac{m}{s}.

(c) We can find the tangential acceleration as follows:


at=αR=3.14 rads20.1 m=0.314 ms2.a_t=\alpha R=3.14\ \dfrac{rad}{s^2}\cdot0.1\ m=0.314\ \dfrac{m}{s^2}.

(d) We can find the centripetal acceleration as follows:


ac=ω2R=(3.14 rads)20.1 m=0.986 ms2.a_c=\omega^2R=(3.14\ \dfrac{rad}{s})^2\cdot0.1\ m=0.986\ \dfrac{m}{s^2}.

(e) We can find the full acceleration from the Pythagorean theorem:


a=ac2+at2=(0.986 ms2)2+(0.314 ms2)2=1.03 ms2.a=\sqrt{a_c^2+a_t^2}=\sqrt{(0.986\ \dfrac{m}{s^2})^2+(0.314\ \dfrac{m}{s^2})^2}=1.03\ \dfrac{m}{s^2}.

(f) We can find the angle as follows:


θ=12αt2,\theta=\dfrac{1}{2}\alpha t^2,θ=123.14 rads2(1 s)2=1.57 rad.\theta=\dfrac{1}{2}\cdot3.14\ \dfrac{rad}{s^2}\cdot(1\ s)^2=1.57\ rad.

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