I don't get the same results in a Pytel dynamics problem

  • Thread starter Thread starter Alexanddros81
  • Start date Start date
  • Tags Tags
    Dynamics Mechancis
Click For Summary

Homework Help Overview

The discussion revolves around a dynamics problem from Pytel Mechanics, specifically problem 12.4, which involves the motion of a bead along a wire described by a quadratic position function. Participants are examining the time at which the bead leaves the wire and the total distance traveled by the bead during its motion.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss their calculations for the distance traveled by the bead, with some arriving at the same result of 800 mm, while questioning the book's provided answer of 2000 mm. Various methods are mentioned, including differentiation to find velocity and graphical analysis to estimate distance.

Discussion Status

The discussion is ongoing, with multiple participants sharing their calculations and reasoning. There is no explicit consensus on the correct distance traveled, but several participants have provided similar results, suggesting a potential common understanding of the problem.

Contextual Notes

Some participants express concerns about the clarity of the problem statement and the dimensions used in the equations, indicating possible ambiguities in the original problem setup.

Alexanddros81
Messages
177
Reaction score
4
Poster has been reminded to post all schoolwork type questions in the Homework Help forums
Hi!
I have Pytel Mechanics Dynamics 2nd edition and revising for my own purposes (not a homework).
I am trying to solve problem 12.4:

A bead moves along a straight 1500-mm wire that lies along the x-axis. The position of the bead
is given by
x=t2 - 10t mm
where x is measured from the center of the wire, and t is the time in seconds. Determine
a) the time when the bead leaves the wire; and b) the distance traveled by the bead from t=0 until
it leaves the wire

I get (a) to be 32.83sec which is correct according to even-numbered answers given at the back of the
book but (b) is given to be 2000mm which is not what I get. I get 800mm

Can you verify if my answer to (b) is correct or tell me what is your answer of (b)?

Does anyone has all the answers of the book I mentioned?

thanks
 
Physics news on Phys.org
I get 800 too: t2 - 10t = (t-5)2 - 25 , so it evaluates to -25 mm at t=5 .

I don't have all the answers :frown:
 
After trying to calculate (b) analytically and failed to get 2000mm
today I differentiated x to get the velocity which is 2t-10. I plotted the graph
and calculated the area of the two triangles. that gave me aproximately 800mm
I used https://www.desmos.com/calculator to do the plot.
 
Alexanddros81 said:
After trying to calculate (b) analytically
It goes 25 mm to the left and then 775 to the right. 800 mm of travel.
Don't waste much more time on this and move on to the next !
 
Also note that this book is very sloppy concerning dimensions. The correct formula must read
$$x(t)=\frac{a}{2} t^2 +v t,$$
where ##a## is the constant acceleration and ##v## the initial velocity. In other words the constants in front of ##t## must have the correct dimensions to be a valid formula in physics.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
7K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 5 ·
Replies
5
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
24
Views
4K