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.
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Last edited by studiot; Aug 20th 2017 at 03:10 PM.
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