# Impulse on a suspended rod.

1. Aug 1, 2012

### AbigailM

For prelim preparation. I've also included a figure of the problem.

1. The problem statement, all variables and given/known data
A stick of length l is suspended by one end at point P so that it hangs vertically and so that the top end of the stick does not move. A horizontal impulse J is applied perpendicular to the stick a distance a below the point of suspension. In general there will be an opposite impulse J' that must be given at the top of the stick (point P) to keep its point of suspension fixed. Find the distance a such that J'=0.

2. Relevant equations
$\tau=r\hspace{1 mm} x\hspace{1 mm} F$

$\tau=I\alpha$

L=T-U

3. The attempt at a solution
$L=\frac{1}{2}I\dot{\theta}^{2} +mg\frac{l}{2}cos\theta$

Using Lagrange's Equations
$I\alpha=-mg\frac{l}{2}sin\theta$

Equating the two equations for torque and setting r=a
$I\alpha = aFsin\theta$

Equate the last to equations
$aFsin\theta=-mg\frac{l}{2}sin\theta$

Solving for F
$F=-\frac{mgl}{2a}$

Therefore J is
$J=-\frac{mglt}{2a}$

So J' is
$J'=\frac{mglt}{2a}$

Is this looking correct? Gracias!

#### Attached Files:

• ###### torque.png
File size:
1.4 KB
Views:
73
Last edited: Aug 1, 2012
2. Aug 1, 2012

### crimsonidol

I don't think that's right.

you have torque T=Fa since it applied perpendicular.
since it gives an impulse to balance the torque. if torque is zero no impulse. then a=0 is the point you are looking for i assume.

Your langrangian should have the constraint term too coming from the applied force.

Last edited: Aug 1, 2012
3. Aug 2, 2012

### rude man

You seemingly haven't solved for a, have you? a(J' = 0)?

I'm not up on Lagrangians any more but might look at the classical solution later.

OK, looked at it, seems simple if I did it right.

If you get an answer I'll tell you if it's what I got.

Hint: torques about the c.g. ?

Last edited: Aug 2, 2012
The text passage contains a physics problem involving the application of a horizontal impulse to a suspended rod and the calculation of the distance from the point of suspension where the impulse should be applied to ensure no opposite impulse is required at the top of the rod. The passage includes a problem statement, relevant equations, and an attempt at a solution, followed by comments from other users.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# Impulse on a suspended rod.

1. Aug 1, 2012

### AbigailM

For prelim preparation. I've also included a figure of the problem.

1. The problem statement, all variables and given/known data
A stick of length l is suspended by one end at point P so that it hangs vertically and so that the top end of the stick does not move. A horizontal impulse J is applied perpendicular to the stick a distance a below the point of suspension. In general there will be an opposite impulse J' that must be given at the top of the stick (point P) to keep its point of suspension fixed. Find the distance a such that J'=0.

2. Relevant equations
$\tau=r\hspace{1 mm} x\hspace{1 mm} F$

$\tau=I\alpha$

L=T-U

3. The attempt at a solution
$L=\frac{1}{2}I\dot{\theta}^{2} +mg\frac{l}{2}cos\theta$

Using Lagrange's Equations
$I\alpha=-mg\frac{l}{2}sin\theta$

Equating the two equations for torque and setting r=a
$I\alpha = aFsin\theta$

Equate the last to equations
$aFsin\theta=-mg\frac{l}{2}sin\theta$

Solving for F
$F=-\frac{mgl}{2a}$

Therefore J is
$J=-\frac{mglt}{2a}$

So J' is
$J'=\frac{mglt}{2a}$

Is this looking correct? Gracias!

#### Attached Files:

• ###### torque.png
File size:
1.4 KB
Views:
73
Last edited: Aug 1, 2012
2. Aug 1, 2012

### crimsonidol

I don't think that's right.

you have torque T=Fa since it applied perpendicular.
since it gives an impulse to balance the torque. if torque is zero no impulse. then a=0 is the point you are looking for i assume.

Your langrangian should have the constraint term too coming from the applied force.

Last edited: Aug 1, 2012
3. Aug 2, 2012

### rude man

You seemingly haven't solved for a, have you? a(J' = 0)?

I'm not up on Lagrangians any more but might look at the classical solution later.

OK, looked at it, seems simple if I did it right.

If you get an answer I'll tell you if it's what I got.

Hint: torques about the c.g. ?

Last edited: Aug 2, 2012
user
The text passage contains a physics problem involving the application of a horizontal impulse to a suspended rod and the calculation of the distance from the point of suspension where the impulse should be applied to ensure no opposite impulse is required at the top of the rod. The passage includes a problem statement, relevant equations, and an attempt at a solution, followed by comments from other users.

TYPE: SINGLE_QUESTION_WITH_ANSWER