Why Does Angular Momentum Calculation Differ When Using Different Formulas?

[Neon32]

Homework Statement

Homework Equations

I= sum m r²
L= r p
or
L=I W

The Attempt at a Solution

I= m1 r1² + m2 r2²
I= 5.20 (0.9)²+ 2.20(0.9)²= 5.994 kg.m²

Then I used the second equation of second momentum
L(Angular momentum) = I W
L= 5.994 x 4.60

In the solutions sheet, he used the first rule: L= r p and he got a different answer than mine: What did I do wrong?

 

[gneill]

Then I used the second equation of second momentum
L(Angular momentum) = I W
L= 5.994 x 4.60

"4.60" is the linear velocity. You need to use an angular velocity for W.

 

[Neon32]

"4.60" is the linear velocity. You need to use an angular velocity for W.

How do I know if it's linear or angular velocity? He didn't mention if it's linear or angular.

 

[gneill]

How do I know if it's linear or angular velocity? He didn't mention if it's linear or angular.

What are the units given for v? What are the units of linear velocity? How about angular velocity?

 

[Neon32]

What are the units given for v? What are the units of linear velocity? How about angular velocity?

Linear velocity has unit m/s
Angular velocit has unit rad/s

I got it :D. Thanks!

 

[Neon32]

What are the units given for v? What are the units of linear velocity? How about angular velocity?

In the second equation L = rxp >> Is cross product between vectors so if we want the magnitude, we should use rxf sin(angle). Why he didn't do that?

 

[gneill]

In the second equation L = rxp >> Is cross product between vectors so if we want the magnitude, we should use rxf sin(angle). Why he didn't do that?

L = r x p is a vector expression. The magnitude of L is given by L = |r||p|sin(θ).

In this instance the angle happens to be θ = 90° . Knowing that sin(90°) = 1 he wrote the simplified expression for the magnitude. Granted, to be technically correct he should have pointed this out in some fashion, but it's a common enough simplification that it shouldn't cause problems interpreting the solution.

 

[Neon32]

L = r x p is a vector expression. The magnitude of L is given by L = |r||p|sin(θ).

In this instance the angle happens to be θ = 90° . Knowing that sin(90°) = 1 he wrote the simplified expression for the magnitude. Granted, to be technically correct he should have pointed this out in some fashion, but it's a common enough simplification that it shouldn't cause problems interpreting the solution.

Can you tell me why the angle between vector r and vecor p is 90? isn't the angle between them =0? since they are in same direction

 

[gneill]

Can you tell me why the angle between vector r and vecor p is 90? isn't the angle between them =0? since they are in same direction

p is a linear momentum of one of the particles. It would be co-linear with the velocity vector of that particle (p = mv). Since the particles are moving in a circle and thus velocities are tangential, the angle between the radius vector and the velocity must be 90°.