Juggler Problem: Solving Kinematics for Ceiling Height

  • Thread starter Thread starter student1ds
  • Start date Start date
Click For Summary

Homework Help Overview

The problem involves a juggler transferring balls between hands and determining the minimum ceiling height required for juggling a specific number of balls. The context is rooted in kinematics, particularly focusing on the motion of the balls as they are tossed and caught.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the time each ball spends in the air and the implications of the 0.20 s transfer time. There are considerations about the timing of tosses and catches, and whether to assume negligible overhead for these actions.

Discussion Status

The discussion is ongoing, with various interpretations of the problem being explored. Some participants suggest that the provided information is sufficient if certain assumptions are made about the timing of tosses and catches. Others are questioning the adequacy of the information regarding the timing dynamics of juggling.

Contextual Notes

Participants are considering assumptions about the timing of catches and throws, particularly whether the 0.20 s transfer time can be treated as the total time between tosses. There is also a mention of the need for common-sense assumptions to facilitate problem-solving.

student1ds
Messages
5
Reaction score
0
A juggler can transfer a ball from his left hand to his right hand in 0.20 s. If he is juggling 5 balls and releases them from a height of 1.5 m, what is the minimum height of the ceiling? How much higher should the ceiling be if he wants to juggle 6 balls ?

How are we going to solve this? If we use kinematics equations, we could problem get through it, but I am not sure how to.
 
Physics news on Phys.org
The first step is to figure out how long each ball spends in the air. (Not the 0.20 s part, but the part where the ball is tossed upward.)
 
Doesn't seem like there's enough information regarding catch and throw times for the juggled balls. If the transfer from left to right hand is a toss (what most jugglers do), then it's possble for the juggler to do the left hand transfer toss before the right hand does the upwards toss, using the .2 second window of transfer toss time to switch between the upward toss and catching the ball from the transfer toss. Same goes for the left hand catching the next ball during the .2 second window transfer toss time. This would reduce the time between tosses. On the other hand, the time between tosses could be greater than the .2 second toss window time.

Perhaps the idea is to assume throw and catch time overhead is effectively zero, so that the time between all tosses is .2 seconds.
 
rcgldr said:
Perhaps the idea is to assume throw and catch time overhead is effectively zero, so that the time between all tosses is .2 seconds.
That is how I am interpreting the problem.
 
All the information required to solve the problem is present, provided one makes some common-sense assumptions, including that 0.20 s is the time from catching a ball with one hand to throwing the same ball with the other. Another assumption would be that he can throw one ball at the same moment he catches the one behind it.

Thus, 0.20 s is simply dwell time at the bottom, as well as the time between each of the balls in flight. The only kinematic equation required is the one for the vertical component.
 
This might help in the visualization of the problem: http://www.flasharcade.com/arcade-games/juggling-simulator-game.html :)
 
Last edited by a moderator:

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
Replies
7
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
2
Views
6K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 4 ·
Replies
4
Views
1K