- #1
MathewsMD
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26. A 2.0-kg block starts from rest on the positive x-axis 3.0 m from the origin and thereafter has an acceleration given by a = (4.0 m/s2)ˆi (3.0 m/s2)ˆj. At the end of 2.0 s its angular momentum about the origin is:
A. 0
B. (36kg·m2/s)kˆ
C. (+48kg·m2/s)kˆ
D. (96kg·m2/s)kˆ
E. (+96kg·m2/s)kˆ
ans: B
For this question, they essentially did:
L = | r x p |
L = (r)(m)(v)
L = (3m)(2kg)(6m/s) * I calculated v = 6 m/s to be the tangential velocity
Now, I was wondering why they didn't change the radius since it accelerates for 2 seconds. During this time period, it moves across the axis, parallel to the original rotation axis.
Δx = (1/2)(4m/s2)(2s)2 and I added this length to my attempt, but it doesn't seem like the solution did. I was wondering if this is correct or if it's incorrect for future reference...
A. 0
B. (36kg·m2/s)kˆ
C. (+48kg·m2/s)kˆ
D. (96kg·m2/s)kˆ
E. (+96kg·m2/s)kˆ
ans: B
For this question, they essentially did:
L = | r x p |
L = (r)(m)(v)
L = (3m)(2kg)(6m/s) * I calculated v = 6 m/s to be the tangential velocity
Now, I was wondering why they didn't change the radius since it accelerates for 2 seconds. During this time period, it moves across the axis, parallel to the original rotation axis.
Δx = (1/2)(4m/s2)(2s)2 and I added this length to my attempt, but it doesn't seem like the solution did. I was wondering if this is correct or if it's incorrect for future reference...