Find an expression for linear momentum and torque

[Stickboy10]

Hi guys, I am hoping you can either point me in the right direction here or show me how to do this a bit. the problem is as follows:

"a particle with weight 86.8 N is positioned at r= (8.1t)i - (7.2t-9.4t²)j.
t is in seconds. Find an expression for angular momentum, L and torque, T which act on this particle." (i is the x vector and j is the y vector)

What I am not getting is how to relate L and T to the weight and position vector. Thanks for any help guys!

 

[ojs]

Do you know how to find angular momentum in general?

Once you know how to find the angular momentum we can talk about the torque.

 

[AlexChandler]

You should know that

$$\vec \tau = \vec r \times \vec F$$

and also that

$$\vec L = \vec r \times m \vec v$$

 

[Stickboy10]

Okay so from this I find that
$$\bar{L}$$ = $$\bar{r}$$ x m$$\bar{v}$$
and
$$\bar{\tau}$$ = $$\bar{r}$$ x m$$\bar{a}$$

The difference between the two is the acceleration from the velocity. Acceleration is the derivative of velocity. So it looks like I find the linear momentum first and then since I have that I can solve for the velocity, take the derivative, sub into the torque equation and then solve the acceleration.

oh... that won't work will it.

 

[AlexChandler]

Okay so from this I find that
$$\bar{L}$$ = $$\bar{r}$$ x m$$\bar{v}$$
and
$$\bar{\tau}$$ = $$\bar{r}$$ x m$$\bar{a}$$

The difference between the two is the acceleration from the velocity. Acceleration is the derivative of velocity. So it looks like I find the linear momentum first and then since I have that I can solve for the velocity, take the derivative, sub into the torque equation and then solve the acceleration.

oh... that won't work will it.

You are given the position vector. You can take a time derivative to find the velocity, and another time derivative to find the acceleration. You have the mass. You have everything. Just plug the position, velocity, and acceleration into the equations and you got it.

remember

$$\vec v = \frac{d \vec r}{dt}$$

$$\vec a = \frac{d \vec v }{dt}$$

 

[Stickboy10]

Oh duh! I don't why I didnt see that! Thank you AlexChandler.