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19 P

Expert-verifiedFound in: Page 335

Book edition
9th Edition

Author(s)
Raymond A. Serway, John W. Jewett

Pages
1624 pages

ISBN
9781133947271

**Question: A wheel 2.00m in diameter lies in a vertical plane and rotates about its central axis with a constant angular acceleration of ${\mathbf{4}}{\mathbf{.}}{\mathbf{00}}{\mathbf{\u200a}}{\mathit{r}}{\mathit{a}}{\mathit{d}}{\mathbf{/}}{{\mathit{s}}}^{{\mathbf{2}}}$. 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, ${\mathit{t}}{\mathbf{=}}{\mathbf{2}}{\mathbf{.}}{\mathbf{00}}{\mathbf{\u200a}}{\mathit{s}}$ 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.**

** a. **The angular speed of the wheel and, for point P is ${\omega}_{f}=8\u200arad/s$.

b. The tangential speed of the wheel is $v=8\u200a\text{m}/\text{s}$.

c. The total acceleration of the wheel is $a=64.1\u200a\text{m}/{\text{s}}^{2}$ at $\text{\theta =3.57\xb0}$

d. The angular position of the wheel is ${\theta}_{f}=9\u200a\text{rad}$.

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.

(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 ${\omega}_{i}=0$, $t=2\u200as$, $\alpha =4\u200a\text{rad}/{\text{s}}^{2}$.

${\omega}_{f}=0+\left(4\right)\left(2\right)=8\u200a\text{rad}/\text{s}$

Therefore, the angular speed of the wheel is ${\omega}_{f}=8\u200a\text{rad}/\text{s}$.

(b)

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

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

Substitute the given values, $r=\frac{d}{2}=\frac{2}{2}=1\u200a\text{m}$, ${\omega}_{f}=8\u200arad/s$

$v=\left(1\right)\left(8\right)=8\u200a\text{m}/\text{s}$

Therefore, the tangential speed is $v=8\u200a\text{m}/\text{s}$.

(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, $r=1\u200am,\alpha =4\u200a\text{rad}/{\text{s}}^{2}$.

${a}_{t}=\left(1\right)\left(4\right)=4\u200a\text{m}/{\text{s}}^{2}$

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

${a}_{r}=\frac{{v}^{2}}{r}$

Substitute the values, $v=8\u200am/s,r=1\u200am$.

${a}_{r}=\frac{{\left(8\right)}^{2}}{1}=64\u200a\text{m}/{\text{s}}^{2}$

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

$\theta ={\mathrm{tan}}^{-1}\left(\frac{{a}_{t}}{{a}_{r}}\right)\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-1}\left(\frac{4}{64}\right)=3.57\xb0$

Therefore, the total acceleration of the wheel is $a=64.1\u200a\text{m}/{\text{s}}^{2}$ at $\theta =3.57\xb0$.

(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 ${\theta}_{i}=0,{\omega}_{i}=0,t=2\u200as,$ $\alpha =4\u200a\text{rad}/{\text{s}}^{2}$.

${\theta}_{f}=1+0+\frac{1}{2}\left(4\right)(2{)}^{2}=9\u200a\text{rad}$

Thus, the angular position of the wheel is ${\theta}_{f}=9\u200a\text{rad}$.

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