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Found in: Page 294

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# A point on the rim of a ${\text{0.75 m}}$-diameter grinding wheel changes speed at a constant rate from ${\text{12 m/s}}$ to${\text{25 m/s}}$ in ${\text{6.2 s}}$. What is the average angular acceleration of the wheel?

Angular acceleration of the wheel is $5.60\text{\hspace{0.17em}\hspace{0.17em}}\text{rad}/{\text{s}}^{2}$.

See the step by step solution

## Step 1: Given

i) Diameter of rim is $0.75\text{\hspace{0.17em}m}$

ii) Initial speed is $12\text{\hspace{0.17em}m}/\text{s}$

iii) Final speed is $25\text{m}/\text{s}$

iv) Time period is $6.2\text{sec}$

## Step 2: Determining the concept

The rate of change of angular velocity of a body with respect to time is called angular acceleration. Find initial as well as final angular velocity using the relationship between rotational and linear velocity. Use a rotational kinematic equation to find the angular acceleration.

The angular velocity is given as-

${\mathbf{v}}{\mathbf{=}}{\mathbf{r\omega }}$

The angular acceleration is given as-

${\mathbf{\alpha }}{\mathbf{=}}\frac{\mathrm{\Delta \omega }}{t}$

where, v is velocity, t is time, r is radius, ${\mathbf{\alpha }}$ is angular acceleration and ${\mathbf{\omega }}$ is angular frequency.

## Step 3: Determining theangular acceleration of the wheel

Here, find the initial as well as final angular speed from the linear velocity as follows:

${v}_{1}=r{\omega }_{1}$

$12\text{\hspace{0.17em}m/s}=\left(0.375\text{\hspace{0.17em}m}\right){\omega }_{1}$

${\omega }_{1}=32\text{\hspace{0.17em}rad/s}$

${v}_{2}=r{\omega }_{2}$

$25\text{\hspace{0.17em}m/s}=\left(0.375\text{\hspace{0.17em}m}\right){\omega }_{2}$

$\omega =66.7\text{\hspace{0.17em}rad/s}$

So, the change in angular speed is as follows:

$\Delta \omega =66.7\text{\hspace{0.17em}rad/s}-32\text{\hspace{0.17em}rad/s}$

$\Delta \omega =34.7\text{\hspace{0.17em}rad/s}$

Now, angular acceleration is as follows:

$\alpha =\frac{\Delta \omega }{t}$

$\alpha =\frac{34.7\text{\hspace{0.17em}rad/s}}{6.2\text{\hspace{0.17em}s}}$

$\alpha =5.60\text{\hspace{0.17em}rad/s}$

Hence, angular acceleration of the wheel is $5.60\text{\hspace{0.17em}\hspace{0.17em}}\text{rad}/{\text{s}}^{2}$.