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21P

Expert-verifiedFound in: Page 293

Book edition
9th Edition

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

Pages
1624 pages

ISBN
9781133947271

**A disk 8.00 cm in radius rotates at a constant rate of 1200 rev / min about its central axis. Determine (a) its angular speed in radians per second, (b) the tangential speed at a point 3.00 cm from its center, (c) the radial acceleration of a point on the rim, and (d) the total distance a point on the rim moves in 2.00 s.**

- The angular speed in radians per second of a disk is ${\omega}_{i}=126\u200arad/s$
- The tangential speed at a pointfrom its center is $v=3.78\u200am/s$.
- The radial acceleration of a point on a rim is ${a}_{r}=1270\u200am/{s}^{2}$ directed towards the centre.
- The total distance a point on the rim moves in 2.00s is $d=20.2\u200am$.

**The tangential speed can be expressed in terms of angular speed, ${\mathit{v}}{\mathbf{=}}{\mathit{\omega}}{\mathit{r}}$ $\omega $**

**is the angular speed, r ****is the radius.**

(a)

To find the angular speed in radians per second, use the conversion factor,

Convert the unit from $(rev/\mathrm{min})to(rad/s):$

${\omega}_{i}=1200\left(\frac{rex}{mix}\right)\left(\frac{2\pi rad}{1rex}\right)\left(\frac{1\mathrm{min}}{60s}\right)\phantom{\rule{0ex}{0ex}}=126\u200arad/s$

Therefore, the angular speed is ${\omega}_{i}=126\u200arad/s$

(b)

The formula for tangential speed is,

$v=\omega r$

Since, the point is in 3.00 cm distance, r = 0.03 m

Substitute the values,

$\omega =126\u200arad/s,\phantom{\rule{0ex}{0ex}}r=0.03\u200am$

Therefore,

$v=\left(126\right)(0.03)=3.78m/s$

Therefore, the tangential speed of a disk at a point 3.00 cm is $v=3.78\u200am/s$.

(c)

Since, the radial acceleration can be expressed in terms of angular speed; the formula is given by,

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

Substitute the values, $\omega =126\u200arad/s,\u200ar=0.08\u200am$

${a}_{r}=(126{)}^{2}(0.08)=1270\u200am/{s}^{2}$

Therefore, the radial acceleration on a rim is ${a}_{r}=1270\u200am/{s}^{2}$ directed towards the centre.

(d)

The formula for the total distance is,

$d=\omega rt$

Substitute the given values,

$\omega =126\u200arad/s,\phantom{\rule{0ex}{0ex}}r=0.08\u200am\phantom{\rule{0ex}{0ex}}t=2\u200as$

Therefore,

$d=(0.08)\left(126\right)\left(2\right)=20.2\u200am$

Thus, the total distance moved by the rim in 2s is d=20.2m.

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