Question

The blades of a ceiling fan have a radius of 0.380 m and are rotating about a fixed axis with an angular velocity of +1.50 rad/s. When the switch on the fan is turned to a higher speed, the blades acquire an angular acceleration of +2.00 rad/s². After 0.500 s has elapsed since the switch was reset, what is (a) the total acceleration (in m/s²) of a point on the tip of a blade and (b) the angle ϕ between the total acceleration → a and the centripetal acceleration → a c ?

Answer 1

Answer:

(A) 4.75 m/s^{2}

(b) 81.5°

Explanation:

radius (r) = 0.38 m

initial angular velocity (ω') = 1.5 rad/s

angular acceleration (α) = 2 rad/s^{2}

time (t) = 0.5 s

(A)   tangential acceleration (a) = α x r

       a = 2 x 0.38 = 0.76 rad/s^{2}

• radial acceleration (a') = $ω^{2} x r$

       final angular velocity (ω) = ω + αt

       ω = 1.5 + (2 x 0.5) = 3.5 m/s

       a' = $3.5^{2} x 0.38$ = 4.7 m/s^{2}

• total acceleration = $\sqrt{a^{2} + a'^{2}}$

       total acceleration = $\sqrt{0.76^{2} + 4.7^{2}}$ = 4.75 m/s^{2}

(B)  tanθ$=\frac{a'}{a}$

      θ= $tan^{-1}\frac{a'}{a}$

      θ= $tan^{-1}\frac{4.7}{0.7}$ = 81.5°