Potential energy (M2)

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Hi guys, Im practising a potential energy question in M2 and my solution to this question is a abit different than what the text book says. Just wanted to make sure im on the right path .
Basically if we assume the right angle triangle with hypotenuse OB then the adjesent to the alpha would be OBCOS(ALPHA) , we also have the height r so the final height is r-obcos(alpha) so the expression for the potential energy would be mg(r-obcos(alpha)) . Attached the text book answer as well as question

The string of the pendulum is inextensible. Hence as we draw the bob from A to B, the distance from O will not change. So OB =?

Reply 2

Alen.m

OP

11

Original post by 16Characters....

The string of the pendulum is inextensible. Hence as we draw the bob from A to B, the distance from O will not change. So OB =?

When the pendulum moves to the point B, its height from the base is less than when it's at pont A maybe the the diagram isnt clear enough coz i drew abit on it but the seconde diagram should be clear enough So OB is basically the hypotenuse of the triangle which has got alpha as an angle. To find the decrease in height we should take away OBcos(alpha) from r but the text book actually used rcos(alpha) and i used OBcos(alpha) im not sure if it makes any difference

Reply 3

16Characters....

13

Original post by Alen.m

When the pendulum moves to the point B, its height from the base is less than when it's at pont A maybe the the diagram isnt clear enough coz i drew abit on it but the seconde diagram should be clear enough So OB is basically the hypotenuse of the triangle which has got alpha as an angle. To find the decrease in height we should take away OBcos(alpha) from r but the text book actually used rcos(alpha) and i used OBcos(alpha) im not sure if it makes any difference