# Solving for Speed & Revolution: F * delta t = p

• Justforthisquestion1
In summary, the conversation is discussing the calculation of the speed and angular velocity of a rod after an initial force is applied. The participants discuss using equations such as F * delta t = p and F * delta t /m = v to determine the speed of the rod's center of mass. They also consider using angular momentum and the moment of inertia to calculate the angular velocity. The conversation ends with a discussion of determining the height and time of flight of the rod.
Justforthisquestion1
Homework Statement
I am looking for the distance x from the center of mass of a rod with I(cm) = 1/12 * m * l^2 were i can apply a Force F over a time delta t so that the rod does excactly one revolution before he is back at its original height.
Relevant Equations
Torque, Rotational Energy, Kinetic Energy, angular momentum
Honestly i have very little idea.
F * delta t = p
F * delta t /m = v
So i know the speed of the rod
And i know that however high the rod is supposed to go, when its back down it should have done excactly one revolution.
I have the feeling that I should
So probably i have to use something like
1/2 *a * T^2 = h
and
omega * T = pi?

Justforthisquestion1 said:
F * delta t = p
F * delta t /m = v
So i know the speed of the rod
[...]
So probably i have to use something like
1/2 *a * T^2 = h
You certainly have enough information that you can know the speed of the rod after the initial force has been applied. With the initial upward velocity in hand, you will be in a position to work on determining the time of flight.

Can you show us the value you arrive at for ##v##?

Will the force of gravity during the initial acceleration affect this in any way?

Note that I approve of an approach where you decide on strategy first -- deciding what you can solve for and what that will then allow you to solve for next. With such a strategy in hand, you can proceed to the grunt work of solving the equations.

Lnewqban
jbriggs444 said:
You certainly have enough information that you can know the speed of the rod after the initial force has been applied.
Did you mean the speed of the rod's center of mass?

To @Justforthisquestion1 : You also have enough information to find the angular speed of the rod as a function of ##x## in terms of F(Δt).

Lnewqban and jbriggs444
I think what a dont understand is, is whether the point of force application has to be included when calculating v or omega. Also the body will translate and rotate which confuses me even more.
So lets say:
F*delta t = m * v
--> v = F * delta t / m
and since i apply F at distance x from cm it should be
omega = F *delta t / (m * x)
Is that correct so far? The force has to cause translation as well as rotation?

Edit no wait i have to use angular momentum dont i?
r * p = I * omega
with p = F * delta t
an I = 1/12 * m * l^2
we get:
omega = 12 * x * F * delta t / (m * l^2)
and then i would get the speed of cm by multiplying omega with l? Or do i use F * delta t = m * v
for v of cm? And why?

Last edited:
Justforthisquestion1 said:
since i apply F at distance x from cm it should be
omega = F *delta t / (m * x)
No, you have to consider the moment of inertia of the rod.
What angular momentum will the force impart to the rod?

Justforthisquestion1
haruspex said:
No, you have to consider the moment of inertia of the rod.
What angular momentum will the force impart to the rod?
Thats kind of a tricky question.
i am using the cm as axis of rotation?
x * p = angular momentum?
-->
x * p = l * omega
with p = F * delta t
an I = 1/12 * m * l^2
we get:
omega = 12 * x * F * delta t / (m * l^2)
and then i would get the speed of cm by multiplying omega with l? Or do i use F * delta t = m * v
for v of cm? And why?

No!
I had it all wrong. So angular momentum and linear momentum are conserved seperately.
So for v and the conservation of linear momentum i can say
p = m * v
--> F * delta t / m = v
But for omega and the conservation of angular momentum i can say
p * x = I * omega
omega = p * x / I

Edit: Okay so now i am stuck again. I now now v and omega.
And as i said before i have a time T in which the stich rises up and falls down again, and does exactly one revolution.
So omega * T = 2 * pi
and 0.5 * m * v^2 = m * g * h
But how to go on?

Last edited:
jbriggs444 said:
You certainly have enough information that you can know the speed of the rod after the initial force has been applied. With the initial upward velocity in hand, you will be in a position to work on determining the time of flight.

Can you show us the value you arrive at for ##v##?

Will the force of gravity during the initial acceleration affect this in any way?

Note that I approve of an approach where you decide on strategy first -- deciding what you can solve for and what that will then allow you to solve for next. With such a strategy in hand, you can proceed to the grunt work of solving the equations.
Gravity is to be neglected during the initial acceleration
For the rest please kindly look to my entry before this one

Justforthisquestion1 said:
omega = 12 * x * F * delta t / (m * l^2)
good
Justforthisquestion1 said:
and then i would get the speed of cm by multiplying omega with l?
No. ω is the rate of rotation about the cm.
You already have the speed of the cm:
Justforthisquestion1 said:
F*delta t = m * v

Justforthisquestion1
haruspex said:
good

No. ω is the rate of rotation about the cm.
You already have the speed of the cm:
Yes because angular momentum and linear momentum are conserved seperately.
Now:
i have a time T in which the rod rises up and falls down again, and does exactly one revolution.
So omega * T = 2 * pi
and 0.5 * m * v^2 = m * g * h
and also v * t - 0.5 * g *T^2 = 2* h
This is because it needs to go up and down so 2 * h
Oh i just realized i can combine these 3! I will be back in a bit when i have calculated it

Last edited:
Does gravity produce torque around the cm? It should not?

Justforthisquestion1 said:
Does gravity produce torque around the cm? It should not?
Right, it does not. In fact, that's almost the definition of cm.

Justforthisquestion1
@Justforthisquestion1

Please see the Latex Math formatting guide. It will help you convey you work, and help others interpret it. It's not laborious to learn the basics.

Justforthisquestion1
erobz said:
@Justforthisquestion1

Please see the Latex Math formatting guide. It will help you convey you work, and help others interpret it. It's not laborious to learn the basics.
Will do

## What does the equation F * delta t = p represent?

The equation F * delta t = p represents the relationship between force (F), the change in time (delta t), and momentum (p). It is derived from the impulse-momentum theorem, which states that the impulse applied to an object is equal to the change in its momentum.

## How is the equation F * delta t = p used in solving for speed?

The equation can be used to solve for speed by first determining the change in momentum (p) of an object. Since momentum is the product of mass (m) and velocity (v), you can rearrange the equation to find the final velocity (speed) after calculating the impulse (F * delta t).

## What units are typically used in the equation F * delta t = p?

In the equation F * delta t = p, force (F) is typically measured in newtons (N), time (delta t) in seconds (s), and momentum (p) in kilogram meters per second (kg·m/s). These units ensure consistency and correctness in calculations.

## Can the equation F * delta t = p be applied to rotational motion?

Yes, the equation can be adapted for rotational motion by considering torque (τ) instead of force (F), and angular momentum (L) instead of linear momentum (p). The analogous equation for rotational motion is τ * delta t = L.

## What are some practical applications of the equation F * delta t = p?

This equation is widely used in various fields such as physics, engineering, and biomechanics. Practical applications include calculating the impact forces in car crashes, determining the forces involved in sports like baseball or football, and analyzing the effects of collisions in particle physics.

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