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?

Answer 1

Answer:

Explanation:

Given

Radius of bicycle wheel $r=0.3\ m$

Initial angular velocity $\omega _0=0$

It rotates 3 revolution in 5 s therefore

$\omega =2\pi 3=\6\pi =18.85\ rad/s$

using $\omega =\omega _0+\alpha t$

where $\alpha =angular\ acceleration$

$\omega =Final\ angular\ velocity$

$t=time$

$\alpha =\frac{18.85}{5}=3.77 rad/s^2$

Total acceleration of any point will be a vector sum of tangential acceleration and centripetal acceleration

$\omega at t=1$

$\omega =0+3.77\times 1=3.77 rad/s$

$a_c=\omega ^2\cdot r$

$a_c=(3.77)^2\cdot 0.3=4.26 m/s^2$

Tangential acceleration $a_t=\alpha \times r$

$a_t=3.77\times 0.3=1.13 m/s^2$

$a_{net}=\sqrt{a_t^2+a_c^2}$

$a_{net}=\sqrt{(1.13)^2+(4.26)^2}$

$a_{net}=4.41 m/s^2$