Answer:
This question is incomplete
Explanation:
This question is incomplete but the missing figure is in the attachment below.
When an object is travelling around a circular path, there is a force that tends to draw that object towards the center of the circular path and keep the object moving in the curved path, that force is called the centripetal force. From this description, it can be deduced that the direction of the centripetal force that acts on the car (in the attachment below) is D.
The name of the force that provides this centripetal force is frictional force. This is the force that prevents the car from slipping off the road; keeping it moving in the curved path.
- 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.
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]