91Ó°ÊÓ

i) Diameter of rim is$0.75\text{″¾}$

ii) Initial speed is$12\text{″¾}/\text{s}$

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

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

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-

$\mathbf{v}\mathbf{=}\mathbf{°ùÓ\neg }$

The angular acceleration is given as-

$\mathbf{Î\pm }\mathbf{=}\frac{\mathrm{ΔÓ\neg }}{t}$

where, v is velocity, t is time, r is radius,$\mathbf{Î\pm }$is angular acceleration and$\mathbf{Ó\neg }$is angular frequency.

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

$v_1=r{\mathrm{Ó\neg }}_1$

$12\text{″¾/s}=\left(0.375\text{″¾}\right){\mathrm{Ó\neg }}_1$

${\mathrm{Ó\neg }}_1=32\text{ r²¹»å/²õ}$

$v_2=r{\mathrm{Ó\neg }}_2$

$25\text{″¾/s}=\left(0.375\text{″¾}\right){\mathrm{Ó\neg }}_2$

$\mathrm{Ó\neg }=66.7\text{ r²¹»å/²õ}$

So, the change in angular speed is as follows:

$\mathrm{Δ}\mathrm{Ó\neg }=66.7\text{ r²¹»å/²õ}−32\text{ r²¹»å/²õ}$

$\mathrm{Δ}\mathrm{Ó\neg }=34.7\text{ r²¹»å/²õ}$

Now, angular acceleration is as follows:

$\mathrm{Î\pm }=\frac{\mathrm{Δ}\mathrm{Ó\neg }}{t}$

$\mathrm{Î\pm }=\frac{34.7\text{ r²¹»å/²õ}}{6.2\text{ s}}$

$\mathrm{Î\pm }=5.60\text{ r²¹»å/²õ}$

Hence, angular acceleration of the wheel is $5.60\text{ â¶Ä‰}\text{rad}/{\text{s}}^2$.