Methods for calculating average velocity

[henry3369]

Homework Statement

A child is pushing a merry-go-round. The angle through which the merry-go-round has turned varies with time according to θ(t)=γt+βt³, where γ= 0.383 rad/s and β= 1.00×10−2rad/s³.

Calculate the average angular velocity ω_av−z for the time interval t=0 to t= 5.50s.

Homework Equations

Given

The Attempt at a Solution

Okay so I solved all the questions above this one which gave me the final and initial angular velocities and I got ω_i = 0.383 rad/s and ω_f = 1.29 rad/s.

When I calculate average angular velocity, why can't I use (ω_i + ω_f)/2
Instead, I have to use (Θ_f - Θ_i)/5.50 to get the correct answer. Shouldn't the first equation yield the same result because an average is the sum of the velocities divided by 2?

 

[Nathanael]

why can't I use (ω_i + ω_f)/2
...
an average is the sum of the velocities divided by 2?

Nooo this is not true at all!

This only applies to one very special case: constant acceleration.

The average speed (over an interval of time) is the constant speed which produces the same displacement (over the same interval). So if you imagine a graph like y(x) or v(t) then the average is the constant (a horizontal line) which produces the same area (over a certain interval of x or t) as the area under the actual curve.

In the special case where y(x) or v(t) has a constant slope (a.k.a. constant acceleration) the average happens to be the midpoint of the line and thus is (y(a)+y(b))/2
(You might like to convince yourself that the midpoint between to points on a graph with a constant slope is the only constant value which gives the same area between the two points.)

I hope I'm not making it confusing, but words tend to do that with visual ideas. It is important to remember that the (y(a)+y(b))/2 or (v_i+v_f)/2 is just a special case which only applies for constant slope (constant acceleration).

 

[henry3369]

Nooo this is not true at all!

This only applies to one very special case: constant acceleration.

The average speed (over an interval of time) is the constant speed which produces the same displacement (over the same interval). So if you imagine a graph like y(x) or v(t) then the average is the constant (a horizontal line) which produces the same area (over a certain interval of x or t) as the area under the actual curve.

In the special case where y(x) or v(t) has a constant slope (a.k.a. constant acceleration) the average happens to be the midpoint of the line and thus is (y(a)+y(b))/2
(You might like to convince yourself that the midpoint between to points on a graph with a constant slope is the only constant value which gives the same area between the two points.)

I hope I'm not making it confusing, but words tend to do that with visual ideas. It is important to remember that the (y(a)+y(b))/2 or (v_i+v_f)/2 is just a special case which only applies for constant slope (constant acceleration).

If the angular acceleration is constant, can (vi+vf)/2 be used?

 

[Nathanael]

If the angular acceleration is constant, can (vi+vf)/2 be used?

Yes. If and ONLY if the angular acceleration is constant.

 

[Brian T]

What do you think is the more general case of avg velocity?

Think: if I went 100 miles in 2 hours, my avg was?