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Torque graph; finding angular velocity |
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| Mar12-07, 10:13 PM | #1 |
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Torque graph; finding angular velocity
An object whose moment of inertia is 4.10 kg/m^2 experiences the torque shown in the graph attached.
What is the object's angular velocity at 2.70 s? Assume it starts from rest. My main problem i think is how to read the graph. How should i interpret the graph? What i would do first is find the angular acceleration by using... aa = angular acceleration t = torque I = moment of inertia aa= t/I I would find the aa for three parts, the part where it is accelerating positively, when it is not increasing, and then when it decreases. I would then use the aa for the equation while putting in the velocity... w = angular velocity T = time w(f) = w(i) + aaT |
| Apr3-08, 01:51 AM | #2 |
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i wish they would reply, i would like help with this also.
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| Apr3-08, 02:00 AM | #3 |
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Well.. you have the Torque as a function of time. Break it up into 3 different functions, one for each of the time interval. At t=2, you can take it to be a part of either of the time interval, just make sure you include it in ONE and ONLY ONE of them.
Once you have [itex]\tau (t)[/itex], divide it by [itex]I[/itex] to get [tex]\alpha (t) = \frac{1}{I} \tau (t)[/tex]. Then integrate it w.r.t to get the angular velocity in each of these intervals using proper limits. This might help you: [tex] \Delta \omega = \int_{t_1}^{t_2} \alpha (t) dt [/tex] Here, [itex]\Delta[/itex] is used to denote that i'm computing the change in angular velocity since the given acceleration was applied. If the initial angular velocity was 0, you can take [itex]\Delta \omega = \omega[/itex]. Once you have done that, you'll need to addup the angular velocities [or their differences rather] attained in each time interval. Do note that, no net torque doesn't mean zero angular velocity, it just means a zero change in angular velocity. Reply if you have any problems with this. |
| Apr3-08, 03:23 AM | #4 |
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Torque graph; finding angular velocity![]() (btw, your original post was at 4am here in London, and just before midnight in New York … if you want a quick answer, you probably need to start a bit earlier!) The torque is only in two parts. (1) The torque increases steadily from t = 0 to 1, according to the formula tau = 2t (so, for example, at t = 1/4, tau = 1/2) (2) The torque is then steady for one second. After that, it disappears suddenly (in no time at all)!
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| Feb23-10, 08:45 AM | #5 |
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Can anyone tell me why does the car engine have maximum power of e.g. 120 horsepower at 4200 revs per minute, and maximum torque of 250 Nm from 1400-2600 revs per minute?
The car seems to have the best acceleration in the range of the greatest torque (1400-2600 revs per minute), but not at maximum power (4200 revs per minute). What is the catch? |
| Feb23-10, 12:38 PM | #6 |
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