Question

A wheel 2.0 m in diameter lies in the vertical plane and rotates about its central axis
2
Take g  10ms
.
with a constant angular acceleration of 4.0rads
t  0 and the radius vector of a point A on the wheel makes an angle of 60o with the
. The wheel starts from rest at horizontal at this instant. Calculate the angular speed of the wheel, the angular

Answer 1

More correctly

A wheel $2 .0$ m in diameter lies in the vertical plane and rotates about its central axis with a constant angular acceleration of $4.0$ $rad/s^2$. The wheel starts from rest at $t=0$ and the radius vector of a point A on the wheel makes an angle of $60°$ 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 $t=2.0$ s.

The angular speed of the wheel

$\omega=\omega_0+\epsilon t=0+4\cdot2=8rad/s$

The angular position of the point A

$\phi=\phi_0+\omega_0t+\frac{\epsilon t^2}{2}=\pi/3+0+\frac{4\cdot2^2}{2}=9.047rad$

$9.047-2\cdot3.14159=2.764rad$ or $\approx158°$ with the horizontal.

The total acceleration

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