An object rotates about a fixed axis, and the angular position of a reference line on the object is given by,$\mathit{\theta }\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{40}{\mathit{e}}^{\mathbf{2}\mathbf{t}}\mathbf{}$where $\mathit{\theta }$is in radians and tis in seconds. Consider a point on the object that is $\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathit{c}\mathit{m}\mathbf{}$from the axis of rotation. At$\mathit{t}\mathbf{=}\mathbf{0}\mathbf{ }\mathbf{\text{s}}$ , what are the magnitudes of the point’s

(a) tangential component of acceleration and

(b) radial component of acceleration?

The angular position of the reference line on the object is, $\theta =0.40e^{2t}$.

The radial distance of the point on the object is, $r=4.0 \text{cm}=4.0\times {10}^{-2} \text{m}$

The spaceship is taking a turn along a circular path. Hence, we need to use the equations that relate the linear variables (v and a) with the corresponding angular variables ( $\omega$and $\alpha$). The two variables are related through the parameter r, the radius of the path.

Formula

$$\begin{gathered} a_t=\alpha r \\ {a}_r=r{\omega }^2 \\ \alpha =\frac{d^2\theta }{d{t}^2} \end{gathered}$$

By using following formula first calculate angular acceleration

$\alpha =\frac{d^2\theta }{d{t}^2}$

It is given that, $\theta =0.40e^{2t}$

$\begin{array}{rcl}\omega & =& \frac{d\theta }{dt}\\ & =& 0.40\left(e^{2t}.\frac{d\left(2t\right)}{dt}\right)\\ \omega & =& 0.40\times 2e^{2t}\\ & & \end{array}$

$\begin{array}{rcl}\omega & =& 0.80e^{2t}\\ \frac{d^2\theta }{d{t}^2}& =& 0.80\left(e^{2t}.\frac{d\left(2t\right)}{dt}\right)\\ & =& 0.80\times 2e^{2t}\\ & =& 1.60e^{2t}\\ & =& 1.60e^{2t} \raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right.\\ & & \\ & & \end{array}$

Now, the tangential component of the acceleration is given as

$a_t=\alpha r$

Substitute all the value in the above equation.

$$\begin{gathered} a_t=\alpha r \\ =\left(1.60e^{2t} \raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right.\right)\times 4.0\times {10}^{-2} \text{m} \end{gathered}$$

$\begin{array}{rcl}At& =& 0 \text{s}\\ \alpha & =& 1.60e^{2t}\\ & =& 1.60\times 1\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right.\\ \alpha & =& 1.60 \raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right.\end{array}$

$\begin{array}{rcl}{a}_t& =& \alpha r\\ & =& 1.60\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}\text{×}4.0\times {10}^{-2} \text{m}}$}\right.\\ & =& 6.4\times {10}^{-2}\raisebox{1ex}{${\displaystyle \text{m}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right.\\ & & \end{array}$

The magnitude of the tangential component of acceleration is $6.4\times {10}^{-2}\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right.$

Step 3: (b) Calculation of radial component of acceleration

The radial component of the acceleration is calculated as

$a_r=r{\omega }^2$

$\begin{array}{rcl}Att& =& 0\text{s},\\ \omega & =& \frac{d\theta }{dt}\\ & =& 0.80e^{2t}\\ & =& 0.80\times 1 \raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\\ \omega & =& 0.80\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\\ & & \\ & & \end{array}$

$a_r=r{\omega }^2$

Substitute all the value in the above equation.

role="math" localid="1660922142356" $\begin{array}{rcl}{a}_r& =& r{\omega }^2\\ & =& \left(4.0\times {10}^{-2}m\right)\times {\left(0.80\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\right)}^2\\ & =& 2.6\times {10}^{-2}\raisebox{1ex}{${\displaystyle \text{m}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right.\end{array}$

The magnitude of the radial component of acceleration is,role="math" localid="1660922172077" $\begin{array}{rcl}& & 2.6\times {10}^{-2}\raisebox{1ex}{${\displaystyle \text{m}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right.\end{array}$ .