What is the angular velocity at t = 2.4 sec based on given graph?

In summary: It is a constant acceleration. Which means Δv/Δt is constant. You can take any points as values.The problem is that you are taking velocity at time of 2.4 secs.
  • #1
SalsaOnMyTaco
32
0

Homework Statement


The graph shows the angular velocity of a rotating wheel as a function of time.
https://online-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/DuPage/Phys1201/summer/homework/Ch-10-Rotation/ang_vel_graph/omega_vs_t-2.jpg

What is the angular displacement of the wheel at t = 2.4 sec?

Homework Equations


ωf=ωi+αt

θ=θ0+ωit+1/2αt2


The Attempt at a Solution


I first try to find the acceleration it takes from initial angular velocity to move to t=2.4s
-6=-9+a(2.4)=1.25
α=1.25

so assuming θ0=0

θ=-9(2.4)+1/2(1.25)(2.4)2
θ=-18?
 
Last edited by a moderator:
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  • #2
any ideas? i also tried W=θ/t
 
  • #3
what about bxh/2= 2.4*-6/2=-7.2 since the θ is the area is between above the line and the horitzonal x
 
  • #4
SalsaOnMyTaco said:

Homework Statement


The graph shows the angular velocity of a rotating wheel as a function of time.
https://online-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/DuPage/Phys1201/summer/homework/Ch-10-Rotation/ang_vel_graph/omega_vs_t-2.jpg

What is the angular displacement of the wheel at t = 2.4 sec?

Homework Equations


ωf=ωi+αt

θ=θ0+ωit+1/2αt2


The Attempt at a Solution


I first try to find the acceleration it takes from initial angular velocity to move to t=2.4s
-6=-9+a(2.4)=1.25
α=1.25

so assuming θ0=0

θ=-9(2.4)+1/2(1.25)(2.4)2
θ=-18?

Your calculated angular acceleration is wrong. The angular velocity at time zero is -9 rad/s, and it is zero at time 6 seconds. So it increases by 9 rad/s over a time interval of 6 seconds.
 
Last edited by a moderator:
  • #5
Chestermiller said:
Your calculated angular acceleration is wrong. The angular velocity at time zero is -9 rad/s, and it is zero at time 6 seconds. So it increases by 9 rad/s over a time interval of 6 seconds.
Thanks,
but shouldn't my final angular velocity be -6 since its asking for the displacemente at 2.4s?
 
  • #6
SalsaOnMyTaco said:
Thanks,
but shouldn't my final angular velocity be -6 since its asking for the displacemente at 2.4s?

It is a constant acceleration. Which means Δv/Δt is constant. You can take any points as values.
The problem is that you are taking velocity at time of 2.4 secs.
You can only get approximate value of v at 2.4 sec from the graph.

Rather than taking value of velocity at 2.4sec, why not use other values that easy and accurate to calculate. It does not matter where you get your Δv and Δt since it is a linear function.
 

Related to What is the angular velocity at t = 2.4 sec based on given graph?

1. What is an Angular Velocity Graph?

An Angular Velocity Graph is a visual representation of the change in angular velocity over time. It shows how quickly an object is rotating at different points in time.

2. How is Angular Velocity calculated?

Angular Velocity is calculated by dividing the change in angle by the change in time. It is measured in radians per second (rad/s) or degrees per second (deg/s).

3. What does the slope of an Angular Velocity Graph represent?

The slope of an Angular Velocity Graph represents the rate of change of angular velocity. A steeper slope indicates a greater change in angular velocity over time.

4. What information can be obtained from an Angular Velocity Graph?

An Angular Velocity Graph can provide information such as the direction and magnitude of angular acceleration, the direction and speed of rotation, and the time it takes for an object to complete one full rotation.

5. How is an Angular Velocity Graph different from a Linear Velocity Graph?

An Angular Velocity Graph shows the change in rotational speed over time, while a Linear Velocity Graph shows the change in linear speed over time. Additionally, the units of measurement for angular velocity and linear velocity are different.

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