Dynamics - Particle kinetics Normal Tangential example 3
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- čas přidán 23. 05. 2020
- Thermodynamics: drive.google.com/file/d/1bFzQ...
Mechanics of Materials: drive.google.com/file/d/1P-pc...
Statics: drive.google.com/file/d/1aYY5...
Dynamics: drive.google.com/file/d/1_JTk...
Why does changing from ds to dØ not change the limits of integration s=ØR and s0=Ø0R
In the summation in the normal direction, why is the W component in the sin(theta) direction?
I would look at the angles at the 3:40 mark in the video.
Keep in mind that for many “block on an incline” problems the figure shows the angle of the incline. So in those problems, the component of weight in the “new” y direction is Wcos(theta). Here, the angle given in the figure is not the angle of the incline, it’s the angle of a line perpendicular to the slope of the incline.
@@engineeringdeciphered Many thanks, I see that now! Brownie points for the speedy reply!
sir. can we find the velocity at theta=60 by using the constant acceleration equations? I mean skater takes 4*sin60 m in the vertical direction and by using constant acc equations ı found the same velocity. Is it true or not?
(A shorter version: To find the velocity at theta=60, can we think of the question as a projectile? )
No, I don’t think constant acceleration equations would work. And no we can’t use projectile equations because of the normal force acting on the object. For projectiles, gravity is the only force.
@@engineeringdeciphered thanks a lot sir
Which is the reasoning behind not inputting angle 60 in that first a^t equation you found (a^t=9.81cos(theta))? Why would inputting the angle immediately give the wrong calculation of a^t?
Because it is not 60 at the beginning. Or the middle. You need to write equations that are true from start to finish. Especially if you are trying to find velocity. If you plugged in 60 degrees too soon, yes, you would get the acceleration at 60 degrees, but that's not what you want. You want the velocity at 60. To get the velocity, you need to use acceleration. And the acceleration is not constant. It changes, so you need to integrate. And you're integrating from start to finish, so don't plug in 60 yet, because it's not true from start to finish. Wait til after you integrate.
Sorry, I think I beat around the bush instead of answering your question. Plugging in 60 at that point WOULD give you the acceleration at that point. But not for any other time. Since you need velocity and that acceleration isn’t constant, you can’t use constant acceleration equations- you have to integrate. And you’re integrating from beginning to end and the acceleration is not that value for the whole integral.
@@engineeringdeciphered thank you! Both replies explain it really well, we want velocity and the acceleration is non-constant so we'll need to integrate it to get the proper velocity we need for that moment