Question

3. [-/1 Points] DETAILS SERPSE10 10.1.P.002. MY NOTES PRACTICE ANOTHER A bar on a hinge starts from rest and rotates with an angular acceleration a = 13 + 7t, where a is in rad/s2 and t is in seconds. Determine the angle in radians through which the bar turns in the first 3.80 s. rad Need Help? Read It Submit Answer

Answer 1

Answer:-  ${\color{DarkRed} \boldsymbol{\theta = 157. 9 \ rad} }$

Explanation:-

Given:

• Angular acceleration of the bar : $\alpha = 13 + 7 t \ \ rad /s^{2}$
• Final time when the bar turns :
  t = 3.80 s
  ​​​​​​​

Relation between angular velocity and angular acceleration is given by,

$\frac{\mathrm{d} \omega }{\mathrm{d} t} = \alpha$

Integrating above equation,

$\int d \omega =\int \alpha \ dt$

$\omega =\int (13 + 7 t) \ dt$

$\omega = 13 t + \frac{7}{2} t^{2} \ \ rad /s$

This is the angular velocity of the bar.

Now, the angle through which the bar turned is given by,

$\theta = \int_{0}^{t} \omega \ dt$

$\theta = \int_{0}^{3.80} \left (13 t + \frac{7}{2} t^{2} \right )\ dt$

$\theta =\left [ \frac{13}{2} t ^{2} + \frac{7}{6} t^{3} \right ]_{0}^{3.80}$

$\theta = \frac{13}{2} (3.80) ^{2} + \frac{7}{6} (3.80) ^{3}$

$\theta = 93.86 + 64.02$

$\boldsymbol{\theta = 157. 9 \ rad}$

This is the angle in radians through which the bar turns.