Question

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

Short answer

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

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$6.2\text{sec}$

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.

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 $5.60\text{rad}/{\text{s}}^2$.