Answer to Question #172496 in Optics for Diocos

Question

A bicycle wheel with radius 0.3 m rotates from rest to 3 rev/s in 5 s. What is the magnitude and direction of the total acceleration vector at the edge of the wheel at 1.0 s?

1) Magnitude - ?

2) Direction - ?

Answer 1

$3\ (rev/s)=6\pi\ (rad/s)$

$\omega=\epsilon t\to\epsilon=\omega/t=6\pi/5=3.77\ (rad/s^2)$

1)

$a=\sqrt{a_n^2+a_\tau^2}=\sqrt{(r\omega^2)^2+(\epsilon r)^2}=\sqrt{(r(\epsilon t)^2)^2+(\epsilon r)^2}=$

$=\sqrt{(0.3\cdot(3.77\cdot1)^2)^2+(3.77\cdot 0.3)^2}=4.41\ (m/s)$ . Answer

2)

$\tan\alpha=\frac{a_n}{a_\tau}=\frac{r\omega^2}{\epsilon r}=\frac{(\epsilon t)^2}{\epsilon}=\epsilon t^2=3.77$

$\alpha \approx75°$relative to the tangent to the wheel . Answer