Angular momentum problem starting from some point on the x-axis.

[raymondmax1]

Homework Statement

A 2.0 kg block is initially at rest on the positive x-axis 2.0 m from the origin. It is acted on by a constant force F = (4.0i − 4.0 j)N. After this force has acted for 3 s, what is the angular momentum of the block (in kg.m2/s) about the origin?

Homework Equations

F=ma
x = (x0) + (v0)t + (1/2)at^2
v = v0 +axt
L=rXP

The Attempt at a Solution

Fx=m(ax) = 2 m/s^2
Fy=m(ay) = -2 m/s^2

(Vx0)=0 (Vx)=0+2(3)=6 m/s
(Vy0)=0 (Vy)=0+(-2)(3)=-6 m/s

so, V= 6i-6j m/s


(rx)=2+1/2(2)(3)^2
=11
(ry)=0+1/2(-2)(3)^2
=-9

so, r= 11i-9 m

L= rXP = rXmV= (11i-6j)X(2)(6i-6j)

= (11i-6j)X(12i-12j)
= (-66k-56k)
=-112k m^2/s

The correct answer is -24k.
I am not sure what I have done wrong.

This is for studying for an exam I have on Saturday.

Thanks,

Ray

 

[tiny-tim]

Welcome to PF!

Hi Ray! Welcome to PF!

A 2.0 kg block is initially at rest on the positive x-axis 2.0 m from the origin. It is acted on by a constant force F = (4.0i − 4.0 j)N. After this force has acted for 3 s, what is the angular momentum of the block (in kg.m2/s) about the origin?
…
F=ma
x = (x0) + (v0)t + (1/2)at^2
v = v0 +axt
L=rXP

uhh?

just use τ = dL/dt ​

 

[raymondmax1]

Thanks!

I'm not sure how to use that formula. I think my prof did it with those other formulas.

do you think you can explain it more deeply??

-Ray :)

 

[raymondmax1]

??

 

[rwisz]

Hi Ray,

Great job on attempting the problem and showing your work. It seems like you have the correct equations and steps, but there are a few small errors in your calculations. Let's take a closer look and see where you went wrong.

In your attempt, you calculated the velocity in the x-direction correctly as 6 m/s. However, in the y-direction, you made a small mistake. The initial velocity in the y-direction should be -6 m/s, not -6i m/s. This is because the block is initially at rest on the x-axis, so there is no velocity in the y-direction. Therefore, the final velocity in the y-direction should also be -6 m/s, not -6i m/s. This will result in a different value for the y-component of the position vector, which will affect your final answer for the angular momentum.

Additionally, when calculating the position vector, you made a small error in the y-component. It should be 2 + (1/2)(-2)(3)^2 = -4, not -9. This is because the block is moving in the negative y-direction, so the displacement in the y-direction should also be negative.

Correcting these errors, your final position vector will be r = 11i - 4j m. Plugging this into the equation for angular momentum, you should get a final answer of -24k m^2/s, which is the correct answer given in the problem.

Keep up the good work and remember to double check your calculations and units to avoid small errors. Good luck on your exam!

Best,