Solving Freely Moving Mass Homework

[dirk_mec1]

Homework Statement

Given: mass m, moment of inertia I and a impulse applied at a distance x form the center of mass (c.m.).

Find the angular speed and the speed of the c.m.

Homework Equations

$$\sum M_O = I \ddot{ \theta}$$$$F\Delta t = \sum m_i v_i$$

The Attempt at a Solution

I can solve for the acceleration but do I have to use a momentum balance to solve for angular speed?

 

[tiny-tim]

Given: mass m, moment of inertia I and a impulse applied at a distance x from the center of mass (c.m.).

Find the angular speed and the speed of the c.m.
…
I can solve for the acceleration but do I have to use a momentum balance to solve for angular speed?

Hi dirk_mec1!

(Do you mean acceleration? It's an impulse … the acceleration is taken to be irrelevant/infinite )

For the angular speed, find the torque of the impulse, then use the angular version of Newton's second law.

 

[dirk_mec1]

Hi dirk_mec1!

Hi tim!

For the angular speed, find the torque of the impulse, then use the angular version of Newton's second law.

Yes, this solves for the acceleration as shown in angular version of Newton second law:

$$\sum M_O = I \ddot{ \theta}$$

But I presume you mean:

$$\sum M_O = I \ddot{ \theta} = I \omega^2 r$$

right?

 

[tiny-tim]

impulse

Hi dirk_mec1!

But I presume you mean:

$$\sum M_O = I \ddot{ \theta} = I \omega^2 r$$

right?

No … wrong formula … that's something like the formula for a continuous centripetal force making the centre of mass move in a large circle.

When the force ends, the circular motion stops!

This is an impulse, an instantaneous force, that makes the body spin on it own axis, with the centre of mass moving in a circle of radius zero.

When the force ends, the circular motion continues!

Newton's second law for impulses takes the forms:

impulse = momentum after minus momentum before;

torque of impulse = angular momentum after minus angular momentum before. ​

 

[dirk_mec1]

Hi dirk_mec1!
impulse = momentum after minus momentum before;

So in formula form:

$$F \Delta t = \Delta p$$

torque of impulse = angular momentum after minus angular momentum before.

$$\mathbf{L}=\sum_i \mathbf{r}_i\times m_i \mathbf{V}_i$$So if I understand you correctly the torque is :

$$F \Delta t \cdot x$$

And how do I proceed?

 

[tiny-tim]

I'm Pulse

Hi dirk_mec1!

(have a delta: ∆)

So in formula form:

$$F \Delta t = \Delta p$$

Yes and no …

F ∆t is correct, if you're given F (a force) and ∆t (a time interval).

But the question doesn't give you those …

it gives you an impulse …

let's call it P (for "I'm Pulse" ) …

Then the formula is P = ∆p.

And the torque is Pxx

$$\mathbf{L}=\sum_i \mathbf{r}_i\times m_i \mathbf{V}_i$$
…
And how do I proceed?

Well, that formula for angular momentum L is correct in principle (though you could have written it L = ∑r_i x p_i) …

but again you're not given the individual r_i and p_i …

you're given the moment of inertia, I …

so L = … ?

and then ∆L = … ?

 

[dirk_mec1]

Hi dirk_mec1!

so L = … ?

Well there´s another formule here in my book it reads:

$$\mathbf{L}= I \mathbf{\omega}$$

and then ∆L = … ?

Pxx?

 

[tiny-tim]

Well there´s another formule here in my book it reads:

$$\mathbf{L}= I \mathbf{\omega}$$

and then ∆L = … ?

Pxx?

Looks good!

 

[dirk_mec1]

So

$$Px=I \omega \rightarrow \omega = \frac{P \cdot x}{ I}$$

right, Tim?