# Question e6799

Feb 22, 2018

See the explanation below

#### Explanation:

Apply the Equations of motion (rotational)

$\omega = {\omega}_{0} + \alpha t$..........$\left(1\right)$

$\Delta \theta = {\omega}_{0} + \frac{1}{2} \alpha {t}^{2}$..............$\left(2\right)$

The initial angular velocity is omega_0=?#

The time is $t = 4 s$

The angle is $\Delta \theta = 162 r a d$

The final angular velocity is $\omega = 108 r a {\mathrm{ds}}^{-} 1$

Substituting those values in equations $\left(1\right)$ and $\left(2\right)$ and solving for ${\omega}_{0}$

$\Delta \theta = {\omega}_{0} + \frac{1}{2} \cdot \frac{\omega - {\omega}_{0}}{2} \cdot {t}^{2}$

$162 = {\omega}_{0} + \frac{1}{4} \cdot \left(108 - {\omega}_{0}\right) \cdot 16$

$162 = {\omega}_{0} + 432 - 16 {\omega}_{0}$

$15 {\omega}_{0} = 432 - 162 = 270$

${\omega}_{0} = \frac{270}{15} = 18 r a {\mathrm{ds}}^{-} 1$

The angular acceleration is

$\alpha = \frac{\omega - {\omega}_{0}}{t} = \frac{108 - 18}{4} = 22.5 r a {\mathrm{ds}}^{-} 2$

The tangential acceleration is

${a}_{\text{tangential}} = r \cdot \alpha = 0.12 \cdot 22.5 = 2.7 m {s}^{-} 2$

The velocity is tangential $v = \omega r$ and the angle is $= {0}^{\circ}$

The total acceleration is

$a = \sqrt{{a}_{T}^{2} + {a}_{C}^{2}}$

The centripetal acceleration is $= {a}_{C}$

And

$\tan \phi = {a}_{T} / {a}_{C}$