Question

A wheel of diameter 40.0 cm starts from rest and rotates with a constant angular acceleration of 3.00 rad/s². Compute the radial acceleration of a point on the rim for the instant the wheel completes its second revolution from the relationship,

(a) ${\mathit{a}}_{\mathbf{r}\mathbf{a}\mathbf{d}}\mathbf{=}{\mathit{\omega }}^{\mathbf{2}}\mathit{r}$and

$\mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{}{\mathbf{a}}_{\mathbf{rad}}\mathbf{=}\frac{{\mathbf{v}}^{\mathbf{2}}}{\mathbf{r}}$

Short answer

Thus, completes the wheel its second revolution from the relationship $16.576\frac{m}{s^2}$.

Thus, Angular velocity at the end of two revolutions is $16.576\frac{m}{s^2}$.

Step-by-step solution

Step:-1 explanation

We know that the formula of angular velocity,

$$\begin{gathered} {\omega }_z^2={\omega }_{z_0}^2+2\alpha \left(\theta -{\theta }_0\right) \\ {\omega }_2=\sqrt{2{\alpha }_z\left(\theta -{\theta }_0\right)} \\ =\sqrt{2\left(\frac{3rad}{s^2}\right)\left(4\mathrm{\pi rad}\right)} \\ =8.68\frac{rad}{s} \end{gathered}$$

Step:-2 concept

$a_{rad}=\frac{v^2}{r}$and we know that $v=r\omega$.

$$\begin{gathered} a_{rad}=\frac{{\left(r\omega \right)}^2}{r} \\ =r{\omega }^2 \end{gathered}$$

Step:-3 calculation

Put the value here,

$$\begin{gathered} 1\frac{rad}{s}=9.549\frac{m}{s^2} \\ 8.68\frac{rad}{s}=82.88 \end{gathered}$$

Hence we get,

$$\begin{gathered} a_{rad}=\left(0.2m\right)\left(82.88\frac{m}{s^2}\right) \\ =16.576\frac{m}{s^2} \end{gathered}$$

Completes the wheel its second revolution from the relationship $16.576\frac{m}{s^2}$.

Step:-4(b)  explanation

We know that $a_{rad}=\frac{v^2}{r}$.

Here we know that,

Here$$\begin{gathered} =v=r\omega \\ {a}_{rad}=\frac{{\left(r\omega \right)}^2}{r} \end{gathered}$$

Step:-5(b) concept

We know that $a_{rad}=\frac{{\left(r\omega \right)}^2}{r}$.

$$\begin{gathered} a_{rad}=\frac{r^2{\omega }^2}{r} \\ {a}_{rad}=r{\omega }^2 \end{gathered}$$

Step:-6(b) Conclusion

Put the value here,

$a_{rad}=\left(0.2m\right)\left(8.68\frac{rad}{s}\right)$

We know that the value of,

$$\begin{gathered} 1\frac{rad}{s}=9.549\frac{m}{s^2} \\ 8.68\frac{rad}{s}=82.88 \end{gathered}$$

Hence we get,

$$\begin{gathered} a_{rad}=\left(0.2m\right)\left(82.88\frac{m}{s^2}\right) \\ =16.576\frac{m}{s^2} \end{gathered}$$

Agular velocity at the end of two revolutions is $16.576\frac{m}{s^2}$.