Using Acceleration to Calculate Gravity

Click For Summary

Homework Help Overview

The discussion revolves around calculating the gravitational field (g) using the time difference (T) between two steel balls released from different heights (X1 and X2). The problem involves kinematic equations and the relationship between distance, time, and acceleration due to gravity.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to derive an equation for g based on the time difference and heights, exploring various kinematic equations. Some participants suggest using initial conditions and rewriting expressions for time in terms of g. Others point out potential errors in simplifications made by the original poster.

Discussion Status

Participants are actively engaging with the problem, providing guidance on how to express time in terms of gravitational acceleration. There is a recognition of mistakes in simplifications, and the conversation is focused on clarifying these points without reaching a definitive conclusion.

Contextual Notes

No additional constraints or imposed homework rules are noted, but the original poster expresses frustration over the complexity of the problem and seeks clarification on their approach.

zx95
Messages
7
Reaction score
0
1. Two steel balls are suspended from electromagnets, and are released simultaneously when the electric current is shut off. They start at heights X1 and X2, with X2 being higher. When they hit the ground, they land T seconds apart. Find an equation for g (the gravitational field) in terms of T, X1 and X2
2. No further information was given.
3. I have about 8 pages of scratch paper and 5 hours of work trying to figure out what is probably something simple. I began by trying to figure out the acceleration from X1 to the ground:

0 = (1/2)at^2 + Vt + X1
-X1 = (1/2)at^2 - Vt
-X1 - Vt = (1/2)at^2
-2(X1+Vt)/t^2 = a

With this equation, I thought I'd be able to somehow figure out the velocity after the ball from X2 height falls from X2 to X1 and then plug that into my result above.

I tried many different things and none seem to match up with the answer my professor is looking for. A nudge in the right direction is all I'm looking for. After so much work, it's really frustrating to be at square 1 still. Thank you in advance for any light you can shed.
 
Physics news on Phys.org
Hi zx95, welcome to PF.
Initial velocities of the steel balls are zero.
So xi = 1/2*g*t1^2. You can rewrite it as
sqrt(2x1/g) = t1.
Similarly find the expression for t2. Then find t2 - t1 which is equal to T.
 
rl.bhat said:
Hi zx95, welcome to PF.
Initial velocities of the steel balls are zero.
So xi = 1/2*g*t1^2. You can rewrite it as
sqrt(2x1/g) = t1.
Similarly find the expression for t2. Then find t2 - t1 which is equal to T.

Thank you for the quick response!

I took this information and came up with the following:

t1 = sqrt(2X1/g) t2 = sqrt(2X2/g)

T = t2 - t1
T = sqrt(2X2/g) - sqrt (2X1/g)
T = sqrt(2(X2-X1)/g)
sqrt(g) = sqrt(2(X2-X1))/T

g = 2(X2-X1)/T^2

I plugged this into the professors software (and looking back in my scratch paper this is the same answer I got through a much more difficult process). Is this answer correct, or am I missing some step that I'm not seeing?
 
T = sqrt(2X2/g) - sqrt (2X1/g)
T = sqrt(2(X2-X1)/g)

The simplification is wrong.
It should be
T = [ sqrt(2x2) - sqrt(2x1)]/sqrt(g). so
sqrt(g) = [ sqrt(2x2) - sqrt(2x1)]/T
 
rl.bhat said:
T = sqrt(2X2/g) - sqrt (2X1/g)
T = sqrt(2(X2-X1)/g)

The simplification is wrong.
It should be
T = [ sqrt(2x2) - sqrt(2x1)]/sqrt(g). so
sqrt(g) = [ sqrt(2x2) - sqrt(2x1)]/T

Thank you very much! I always make mistakes like that. I need to work on that. Thanks!
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
Replies
12
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
1
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K