contestada

He angular position of a swinging door is described by θ = 5.00 + 10.0 t + 2.00 t2 [rad]. (a) determine the angular position, angular speed, and angular acceleration of the door at t = 3.00 s. (b) if the door starts from rest and its angular acceleration is 0.560 rad/s2, through what angle does it have to turn to reach an angular speed of 0.750 rad/s?

Respuesta :

skyluke89
(a) The angular position of the door is described by
[tex]\theta(t)=5+10t+2t^2 [rad][/tex]

The angular velocity is given by the derivative of the angular position:
[tex]\omega(t)=10+4t [rad/s][/tex]

While the angular acceleration is given by the derivative of the angular velocity:
[tex]\alpha(t)=4 [rad/s^2][/tex]

We want to find the values of these quantities at time t=3.00 s, so we must substitute t=3.00 s into the expressions for [tex]\theta, \omega, \alpha[/tex]:
[tex]\theta(3.00 s)=5+(10)(3.00 s)+2(3.00s)^2 = 53 rad[/tex]
[tex]\omega(3.00 s)=10+4(3.00s)=22 rad/s[/tex]
[tex]\alpha(3.00s)=4 rad/s^2[/tex]

(b) The door starts from rest, so its initial angular velocity is [tex]\omega_i=0 rad/s[/tex], and it reaches a final angular velocity of [tex]\omega_f=0.750 rad/s[/tex] with an angular acceleration of [tex]\alpha=0.560 rad/s^2[/tex]. We can find the angular distance covered by the door by using the following relationship:
[tex]2 \alpha \theta = \omega_f^2 - \omega_i^2[/tex]
from which we find
[tex]\theta= \frac{\omega_f^2}{2 \alpha}= \frac{(0.750 rad/s)^2}{2 \cdot 0.560 rad/s^2} =0.502 rad [/tex]
leena

Hi there!

We are given that:

[tex]\theta = 5 + 10t + 2t^2[/tex]

The equation for angular velocity is equivalent to the derivative of this equation, so:

[tex]\omega(t) = \frac{d\theta}{dt} = 10 + 4t[/tex]

Angular acceleration is the derivative of the function for angular velocity:

[tex]\alpha(t) = \frac{d\omega}{dt} = 4[/tex]

At t = 3 sec:

[tex]\theta = 5 + 10(3) + 2(3^2) = \boxed{53 rad}[/tex]

[tex]\omega = 10 + 4(3) = \boxed{22 rad/sec}[/tex]

[tex]\alpha = \boxed{4 rad/sec^2}[/tex]

We can use the rotational equivalent of a kinematic equation to solve:

[tex]\omega_f^2 = \omega_i^2 + 2\alpha\theta[/tex]

Plug in the givens:

[tex].750^2 = 0^2 + 2(.56)\theta\\\\\frac{.750^2}{2(.56)} = \theta = \boxed{0.502 rad}[/tex]

Otras preguntas