• Support PF! Buy your school textbooks, materials and every day products via PF Here!

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

1. The problem statement, all variables and given/known data

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?


2. Relevant equations

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


3. 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

Science Advisor
Homework Helper
25,790
242
Welcome to PF!

Hi Ray! Welcome to PF! :smile:
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? :confused:

just use τ = dL/dt :wink:
 
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 :)
 
???????????
 

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top