class3 - From last class: v = rr + (r). (42) By...

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From last class: v r ˆ r +( r ˙ θ ) ˆ θ. (42) By diferentiating this with respect to t , we can also show that a = (¨ r - r ˙ θ 2 r r ¨ θ + 2 ˙ r ˙ θ ) ˆ θ (43) Circular motion : Consider the general case o± a particle moving around a circle with center 0 and radius b , not necessarily with constant speed. Since r = b , then ˙ r r = 0, and the velocity o± a particle is v =( b ˙ θ ) ˆ θ (44) The ±ormula ±or the acceleration becomes a = - b ˙ θ 2 ˆ r + b ¨ θ ˆ θ (45) also remember that the circum±erential velocity v = b ˙ θ . example: The bob o± a pendulum moves on a vertical circle o± radius b , and when the string makes an angle θ with the downward vertical, the velocity o± the bob is v 2 =2 gb cos θ (46) where g is a positive constant. Find the acceleration o± the bob when the string is at an angle θ . solution: From the acceleration ±ormula: a = - v 2 b ˆ r v ˆ θ = - (2 g cos θ r v ˆ θ (47) we must diferentiate to get ˙ v : d dt v 2 v ˙ v = - (2 sin θ ) ˙ θ (48) and since v = b ˙ θ , ˙ v = - g sin θ (49)
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class3 - From last class: v = rr + (r). (42) By...

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