# Tough physics pendulum problem, can you help?

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Orangutans can move by brachiation, swinging like a pendulum beneath successive handholds. If an orangutan has arms that are 0.90 m long and repeatedly swings to a 20 degree angle, taking one swing immediately after the other, estimate how fast it is moving in m/s?

• we should really be treating the orangutan as a physical pendulum not a simple pendulum, but there is not enough information given to do so, so it seems we will have to treat them as a simple pendulum

we can calculate easily the period of such a pendulum from

P = 2 pi Sqrt[L/g] = 6.28 Sqrt[0.9m/9.8m.s.s] = 1.90s

the time from the lowest point to the next branch is 1/4 of a period, so this time is 0.48s

…if they mean forward horizontal velocity, draw a triangle representing the situation to see that the horizontal distance covered is given by

hor distance = L sin 20 = 0.9 sin 20 = 0.31m, and the forward speed is

0.31m/0.48s = 0.65m/s

• Take the momentum of the Bullet P=m*v Block momentum after the bullet hits it incredibly is going to be the same (momentum conservation) so which you have M*V=m*v, greater precisely it incredibly is (M+m)*V=m*v with the aid of fact the bullet now’s a element of the block. right here V is the Block with bullet speed. That leads you to V=m*v/(M+m) and kinetic power of this block & bullet is okay=(M+m)*squareV) as they pass upwards that Kinetic power turns into skill power and finally while it stops each and all the kinetic power has switched over to skill power whose formula is U=(M+m)*g*h (g is gravity and h the peak). So in basic terms equate ok=U protecting in concepts what V is from momentum conservation after which you will sparkling up for height. Dont be affraid of letters they’re in basic terms a container with some quantity interior. That makes them simpler than numbers themselves.. rigidity interior the string could be calculated with: T-(M+m)*g=(m+M)*squareV)/l it incredibly is Newtons’ 2d regulation utilized to around flow

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