Question

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 at 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

Answer on Question #72594 Physics / Mechanics | Relativity

Wheel $R = 2.0 \, \text{m}$ in diameter lies in the vertical plane and rotates about its central axis with a constant angular acceleration of $= 4.0 \, \text{rad} \cdot \text{s}^{-2}$. The wheel starts at rest at $t = 0$ and the radius vector of a point A on the wheel makes an angle of $60{}^{\circ} = \pi / 3$ 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 $t = 2.0 \, \text{s}$.

Solution:

The equation of motion for the point A

$$\varphi = \varphi_0 + \omega_0 t + \frac{\varepsilon t^2}{2} = \frac{\pi}{3} + 0 t + \frac{4.0 t^2}{2} = \frac{\pi}{3} + 2 t^2$$

The angular speed of the wheel

$$\omega = \omega_0 + \varepsilon t = 4 t$$

Total acceleration

$$a = \sqrt{\varepsilon^2 + (\omega^2 R)^2}$$

At the instant $t = 2.0$

$$\varphi = \frac{\pi}{3} + 2 \times 2^2 = \frac{\pi}{3} + 2 \times 2^2 = 9.05 \, \text{rad} = 518{}^{\circ}$$ $$\omega = 4 \times 2 = 8 \frac{\text{rad}}{\text{s}}$$ $$a = \sqrt{4^2 + (8^2 \times 2)^2} = 128 \, \text{m/s}^2$$

Answers:

$518{}^{\circ}$

$$8 \frac{\text{rad}}{\text{s}}$$ $$128 \, \text{m/s}^2$$

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