Question

A wheel 2.0 m in diameter lies in the vertical plane and rotates about its central axis
with a constant angular acceleration of 4.0rads .
2
The wheel starts from rest at
t  0 and the radius vector of a point A on the wheel makes an angle of 60º with the
horizontal at this instant. Calculate the angular speed of the wheel, the angular
position of the point A and the total acceleration at t  2.0s.

Answer 1

1) The angular speed of the wheel

$\omega=\omega_0+\epsilon t=0+\epsilon t=\epsilon t=4\cdot 2=8 rad/s$

2) The angular position of the point A

$\phi=\phi_0+\omega_0+\frac{\epsilon t^2}{2}=\frac{\pi}{3}+0+\frac{4\cdot 2^2}{2}=1.047+8=9.047 rad$

or

$\phi=518.4°$

$518.4°-360°=158.4°$with the horizontal

3) The total acceleration

$a=\sqrt{a^2_n+a^2_\tau}=\sqrt{(\frac{v^2}{R})^2+\epsilon^2R^2}=\sqrt{(\frac{\omega^2\cdot R^2}{R})^2+\epsilon^2R^2}=\sqrt{(\omega^2\cdot R)^2+\epsilon^2R^2}==\sqrt{(8^2\cdot 1)^2+4^2\cdot 1^2}=64.1m/s^2$