Question

Question: A wheel 2.00m in diameter lies in a vertical plane and rotates about its central axis with a constant angular acceleration of $\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{ }\mathit{r}\mathit{a}\mathit{d}\mathbf{/}{\mathit{s}}^{\mathbf{2}}$. The wheel starts at rest at t=0, and the radius vector of a certain point on the rim makes an angle of$\mathbf{57}\mathbf{.}{\mathbf{3}}^{\mathbf{o}}$ with the horizontal at this time. At,$\mathit{t}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{ }\mathit{s}$ find (a) the angular speed of the wheel and, for point P, (b) the tangential speed, (c) the total acceleration, and (d) the angular position.

Short answer

a.The angular speed of the wheel and, for point P is ${\omega }_f=8 rad/s$.

b. The tangential speed of the wheel is $v=8 \text{m}/\text{s}$.

c. The total acceleration of the wheel is$a=64.1 \text{m}/{\text{s}}^2$at$\text{θ=3.57°}$

d. The angular position of the wheel is ${\theta }_f=9 \text{rad}$.

Step-by-step solution

Deriving the angular speed formula

The formula that is used to find the angular speed is,${\omega }_f={\omega }_i+\alpha t$

where${\omega }_i$ is the initial position,$\alpha$ is the angular acceleration, t is the time.

Deriving the angular speed of the wheel

(a)

Hence, model the wheel as a rigid object under constant angular acceleration.

Use the formula to find the final angular speed at point P,

${\omega }_f={\omega }_i+\alpha t$

Substitute the given values ${\omega }_i=0$, $t=2 s$, $\alpha =4 \text{rad}/{\text{s}}^2$.

${\omega }_f=0+\left(4\right)\left(2\right)=8 \text{rad}/\text{s}$

Therefore, the angular speed of the wheel is ${\omega }_f=8 \text{rad}/\text{s}$.

Determination of the tangential speed

(b)

Since the tangential speed can be expressed in terms of angular speed, the formula is

$v={\omega }_{f}r$

Substitute the given values, $r=\frac{d}{2}=\frac{2}{2}=1 \text{m}$,${\omega }_f=8 rad/s$

$v=\left(1\right)\left(8\right)=8 \text{m}/\text{s}$

Therefore, the tangential speed is $v=8 \text{m}/\text{s}$.

Calculation of the total acceleration of the wheel

(c)

The total acceleration is the sum of the tangential acceleration and radial acceleration that are perpendicular to each other.

Find the tangential acceleration of the wheel at that can be expressed in terms of angular acceleration.

$a_t=r\alpha$

Substitute the values, $r=1 m,\alpha =4 \text{rad}/{\text{s}}^2$.

$a_t=\left(1\right)\left(4\right)=4 \text{m}/{\text{s}}^2$

Determine the radial acceleration of the wheel that are expressed in terms of linear speed.

$a_r=\frac{v^2}{r}$

Substitute the values, $v=8 m/s,r=1 m$.

$a_r=\frac{{\left(8\right)}^2}{1}=64 \text{m}/{\text{s}}^2$

Thus, the total acceleration can be found using the Pythagorean theorem, since both tangential and radial are perpendicular.

$a=\sqrt{a_t^2+a_r^2}=\sqrt{4^2+{64}^2}=64.1\text{m}/{\text{s}}^2$

Hence, it makes an angle of radial acceleration vector of

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{a_t}{a_r}\right) \\ ={\tan }^{-1}\left(\frac{4}{64}\right)=3.57° \end{gathered}$$

Therefore, the total acceleration of the wheel is $a=64.1 \text{m}/{\text{s}}^2$at $\theta =3.57°$.

Obtaining the angular position

(d)

For the rigid object under constant angular acceleration, the equation that is used to find angular position is

${\theta }_f={\theta }_i+{\omega }_{i}t+\frac{1}{2}\alpha t^2$

Substitute the values ${\theta }_i=0,{\omega }_i=0,t=2 s,$$\alpha =4 \text{rad}/{\text{s}}^2$.

${\theta }_f=1+0+\frac{1}{2}\left(4\right)(2{)}^2=9 \text{rad}$

Thus, the angular position of the wheel is ${\theta }_f=9 \text{rad}$.