Physics Help Forum d'alembert worked example but is trig correct?

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 Aug 20th 2017, 01:45 PM #1 Senior Member   Join Date: Nov 2013 Location: New Zealand Posts: 519 d'alembert worked example but is trig correct? Trying to follow another D'Alembert problem (see attached). Can't figure the step where they wrote $\displaystyle \delta W_G = -mgR \sin \phi \delta \phi$, shouldn't it be $\displaystyle \delta W_G = -mgR (1 - \cos \phi) \delta \phi$ since we are measuring against the vertical force of gravity? Attached Thumbnails     Last edited by kiwiheretic; Aug 20th 2017 at 02:10 PM. Reason: mistake in formula
 Aug 20th 2017, 03:54 PM #2 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 688 Yes, the book is right. Here is the outline derivation. In the attachment Fig 1 show the tangent at A rotated through an angle theta at B. This is the same as the angle turned through by the radius to get from A to B and is completely general. You should be able to convince yourself of this. Now let the radius arm move on a small extra delta theta to C as in Fig 2 we generate a small triangle BCH with hypotenuse aligned to the tangent at B, which we know is at angle theta to the horizontal (tangent at A). We are only interested in the extra work, delta W, created in moving from B to C which is given by the rise delta y times mg, as you corectly observe (with appropriate signs). Now approximate hypotenuse BC by arc BC which is equal to radius times angle. So the rise, delta y, is BC sine theta, which is R delta theta sine theta, as required. Sorry I used theta instead of phi. Attached Thumbnails   Last edited by studiot; Aug 20th 2017 at 04:10 PM.
 Aug 20th 2017, 04:48 PM #3 Senior Member   Join Date: Nov 2013 Location: New Zealand Posts: 519 But what is the length of the red segment (see attached)? I'm not sure how moving that angle around to the tangent helps. Edit: Ok, figured that part out. The red segment is $\displaystyle R\tan \frac {\theta}{2}$ so the work against gravity is $\displaystyle \delta W_g = -mgR\tan \frac {\theta}{2} \sin \theta \: \delta x$ ?? I have a feeling I'm not tied into the right differential $\displaystyle \delta s$ but not sure how to make the substitution. Edit again: Ok, I see what you are getting at now. I was trying to calculate the entire height y rather than the incremental height $\displaystyle \delta y$. Attached Thumbnails     Last edited by kiwiheretic; Aug 20th 2017 at 07:34 PM. Reason: fix up formula

 Tags correct, dalembert, trig, worked

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