Rotational motion- steam engine

AI Thread Summary
The discussion focuses on calculating the total angle a steam engine's flywheel turns after 47.5 seconds, starting from rest with a constant angular acceleration of 1.35 rad/s² for the first 33.1 seconds. Participants emphasize the importance of separating the motion into two phases: the acceleration phase and the constant velocity phase. The correct approach involves using the equations for angular displacement during both phases and summing the results. A key point raised is the need to ensure the final answer is in radians, not degrees, which was a common source of error. Ultimately, the correct method leads to the right solution when applied properly.
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Homework Statement



The flywheel of a steam engine begins to rotate from rest with a constant angular acceleration of 1.35 rad/s^2. It accelerates for 33.1 s, then maintains a constant angular velocity. Calculate the total angle through which the wheel has turned 47.5 s after it begins rotating.

Homework Equations



1. w=wi +alpha(time)

2. theta - theta(initial) = 1/2(w +wi)t

The Attempt at a Solution


I have attempted this problem several times and keep getting the answer wrong! (note: I do not know the correct answer for this particular problem)

First I used equation 1 to find the constant angular velocity. Then, since I know that constant velocity means zero acceleration, I used equation 2 to find the total angle. What am I doing wrong? Please help!:confused:
 
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fruitl00p said:

Homework Statement



The flywheel of a steam engine begins to rotate from rest with a constant angular acceleration of 1.35 rad/s^2. It accelerates for 33.1 s, then maintains a constant angular velocity. Calculate the total angle through which the wheel has turned 47.5 s after it begins rotating.

Homework Equations



1. w=wi +alpha(time)

2. theta - theta(initial) = 1/2(w +wi)t
It would help to graph the angular speed as a function of time. The area under the graph is what you are trying to calculate.

If it starts at \omega = 0 then \omega = \alpha t and:

\theta_1 = \frac{1}{2}\alpha t_1^2 (angle after 33.1 seconds)

\theta_2 = \alpha(t_1) (t_f - t_1) (increase in angle to t= 47.5 seconds)

which is essentially what you have already figured out. Just add those two equations to get \theta = \theta_1 + \theta_2 and solve.

AM
 
There are 2 phases - 33.1s of acceleration and 14.4s of constant velocity.
You must calculate the number of turns for each phase and add them.
Adapt the well known relations

distance = 1/2*acc*time^2
distance = velocity*time

to the angular case.
 
thank you very much, I got the answer correctly. :smile:
 
I am actually using θ1= ½ αt12 and θ2=α (t1)(tf-t1) and adding both angles together but I am not getting the answer right??

thanks
 
Last edited:
aha, the answer was in radians, not degrees
 
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