- During a certain period, the angular position of a rotating object is given by: = − + , where  is in radian and t is in seconds. Determine the angular position, angular speed, and angular acceleration of the rotating object at = Sec.

Answers

Answer 1

The question is not complete. The complete question is :

During a certain period of time, the angular position of a rotating object is given by [tex]$\theta =2t^2 +10t+5$[/tex], where θ is in radians and t is in seconds.  Determine the angular position, angular speed, and angular acceleration of the door (a) at t = 0.00 seconds, (b) at t = 3.00 seconds.

Solution :

Given :

Displacement or angular position of the object, [tex]$\theta =2t^2 +10t+5$[/tex]

∴ Angular speed is   [tex]$\omega = \frac{d \theta}{dt}$[/tex]

                                 ω = 10 + 4t

And angular acceleration is [tex]$\alpha = \frac{d \omega}{dt}$[/tex]

                                              α = 4

a). At time, t = 0.00 seconds :

   Angular displacement is [tex]$\theta =2t^2 +10t+5$[/tex]

                                            [tex]$\theta =2(0)^2 +10(0)+5$[/tex]

                                               = 5 rad

  Angular speed is  ω = 10 + 4t

                                 ω = 10 + 4(0)

                                     = 10 rad/s

Angular acceleration is α = 4 [tex]$rad/s^2$[/tex]

b). At time, t = 3.00 seconds :

   Angular displacement is [tex]$\theta =2t^2 +10t+5$[/tex]

                                            [tex]$\theta =2(3)^2 +10(3)+5$[/tex]

                                               = 53 rad

  Angular speed is  ω = 10 + 4t

                                 ω = 10 + 4(3)

                                     = 22 rad/s

Angular acceleration is α = 4 [tex]$rad/s^2$[/tex]

                                   

                                           


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