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

AI Thread Summary
The discussion focuses on calculating the angular displacement of a wheel at t = 2.4 seconds based on a provided graph of angular velocity over time. Participants clarify that the initial angular velocity is -9 rad/s and increases to 0 rad/s at 6 seconds, indicating a constant angular acceleration. The initial attempt to calculate angular displacement using incorrect acceleration values is corrected, emphasizing the need to use accurate data points from the graph. It is suggested to derive angular velocity at 2.4 seconds from the graph for better accuracy rather than relying solely on calculations. The conversation highlights the importance of understanding linear functions in motion analysis.
SalsaOnMyTaco
Messages
32
Reaction score
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:
Physics news on Phys.org
any ideas? i also tried W=θ/t
 
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
 
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:
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?
 
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.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top