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### Physics For Scientists & Engineers

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.

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

## Step 1: The tangential speed formula

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

is the angular speed, r is the radius.

## Step 2: Calculation of the angular speed of a disk

(a)

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

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

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

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

## Step 3: The tangential speed

(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,

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

Therefore,

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

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

## Step 4: Determination of the radial acceleration on a rim

(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, $\omega =126 rad/s, r=0.08 m$

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

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

## Step 5: Calculation of the total distance moved by the rim in 2s:

(d)

The formula for the total distance is,

$d=\omega rt$

Substitute the given values,

$\omega =126 rad/s,\phantom{\rule{0ex}{0ex}}r=0.08 m\phantom{\rule{0ex}{0ex}}t=2 s$

Therefore,

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

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