Halliday Resnick Krane Chapter 2 Problem 30: Juggling Five Balls

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To solve the juggling problem, it's essential to determine the time each ball is in the air, which is approximately 1.28 seconds for a height of two meters. The challenge arises when juggling multiple balls released at different times, complicating the calculation of touch events. A systematic approach involves counting the number of times each ball touches a hand per minute. Understanding the timing and sequence of each ball's release is crucial for accurate calculations. Overall, careful analysis of the juggling pattern will yield the total number of touch events.
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Homework Statement
A juggler juggles 5 balls with two hands. Each ball rises 2 meters above her hands. Approximately how many times per minute does each hand toss a ball?
Relevant Equations
y= y0 + vt + (at^2/2)
If each ball rises two meters, then the time is in the air is equal to around 1.28 seconds. After this, what am I supposed to do? I feel really stuck. If all of the balls are released at different times, how am I supposed to find the answer?
 
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What will happen if you don't handle each of the other balls before you re-handle the first ball?
 
.. or start with, how many times a minute does a given ball touch a hand? How many touch events are there in total per minute?
 
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vibha_ganji said:
If all of the balls are released at different times, how am I supposed to find the answer?
By counting?
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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