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 4.0 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 73280, Physics / Mechanics | Relativity Question

A wheel $2.0\mathrm{m}$ 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 $60{}^{\circ}$ 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 $t = 2.0\mathrm{s}$.

**Solution.** The equation that gives the angular speed $\omega(t)$ of the wheel is

$$\omega(t) = \omega_0 + \alpha t$$

where $\omega_0$ is the initial angular speed, $\omega_0 = 0$, $\alpha$ is the angular acceleration, $\alpha = 4.0 \, \text{rad/s}^2$. Substituting knowing numbers, we find the angular speed of the wheel at $t = 2.0\,\text{s}$.

$$\omega(t = 2\,\text{s}) = 0 + 4.0 \, \frac{\text{rad}}{\text{s}^2} \cdot 2.0\,\text{s} = 8.0 \, \frac{\text{rad}}{\text{s}}$$

The equation that gives the angular position of the point A is

$$\theta(t) = \theta_0 + \omega_0 t + \frac{1}{2} \alpha t^2$$

where $\theta_0 = 60{}^\circ$ is the initial angular position of the radius vector of a point A. Substituting knowing numbers, we find the angular position of the point A at $t = 2.0\,\text{s}$

$$\theta(t = 2) = 60{}^\circ + 0 + \frac{1}{2} \cdot 4.0 \, \frac{\text{rad}}{\text{s}^2} \cdot (2.0\,\text{s})^2 = 60{}^\circ + 8.0\,\text{rad}$$

Define how many degrees are equal to 8 radians. It is known that $1\,\text{rad} = 57.2958{}^\circ$, then

$$8\,\text{rad} = 8 \cdot 57.2958{}^\circ \approx 458.0{}^\circ$$

and

$$\theta(t = 2\,\text{s}) \approx 60{}^\circ + 458.0{}^\circ = 518.0{}^\circ$$

In order to find the total acceleration, we first find the tangential $a_\tau$ and normal $a_n$ acceleration at $t = 2.0\,\text{s}$

$$a_\tau(t = 2\,\text{s}) = \alpha r = 4.0 \, \frac{\text{rad}}{\text{s}^2} \cdot 1.0\,\text{m} = 4.0 \, \frac{\text{m}}{\text{s}^2}$$ $$a_n(t = 2\,\text{s}) = r \omega^2 = 1.0\,\text{m} \cdot \left(8.0 \, \frac{\text{rad}}{\text{s}}\right)^2 = 64.0 \, \frac{\text{m}}{\text{s}^2}$$

The equation that gives the total acceleration is

$$a_{tot} = \sqrt{a_\tau^2 + a_n^2}$$

Substituting knowing numbers, we find

$$a_{tot}(t = 2\,\text{s}) = \sqrt{(4.0)^2 + (64.0)^2} = \sqrt{16 + 4096} \approx 64.125 \, \frac{\text{m}}{\text{s}^2}$$

**Answer:**

$$\omega(t = 2\,\text{s}) = 8.0 \, \frac{\text{rad}}{\text{s}}$$ $$\theta(t = 2\,\text{s}) \approx 518.0{}^\circ$$ $$a_{tot}(t = 2\,\text{s}) \approx 64.125 \, \frac{\text{m}}{\text{s}^2}$$

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