Finding the magnitude of an object's angular momentum about the origin.

[penguinnnnnx5]

Homework Statement

A 1.4 kg object at x = 2.00 m, y = 3.10 m moves at 4.9 m/s at an angle 45° north of east. Calculate the magnitude of the object's angular momentum about the origin.

Answer is 5.3 kg m^2 /s

Homework Equations

These were the only three I could think of:
L=Iω
L=mvr
ω=v/r

The Attempt at a Solution

First attempt:

L=mvr

L= (1.4)(4.9)sqrt(4+3.1^2)

L=25.3077, not the right answer

Second attempt:

Diagram here.
I tried to find the tangential velocity as shown:

∠2 = arctan (3.10/2.00) = 57.17°

Vtan = 4.9cos(45) / cos(57.17) = 6.390914 m/s

ω=v/r=6.390914/Sqrt(4+3.1^2) = 1.73234 rad/s

Then I'm not really sure how to proceed after this because I'm not entirely sure what equation to use for I.

Third attempt:

Vtan=v ∴

L=1.4(6.390914)(sqrt(4+3.1^2) = 8.94728 m/s, not the right answer

Is my diagram wrong or am I missing something very important?

 

[ehild]

Homework Statement

A 1.4 kg object at x = 2.00 m, y = 3.10 m moves at 4.9 m/s at an angle 45° north of east. Calculate the magnitude of the object's angular momentum about the origin.

Answer is 5.3 kg m^2 /s

Homework Equations

These were the only three I could think of:
L=Iω
L=mvr
ω=v/r

The Attempt at a Solution

First attempt:

L=mvr

L= (1.4)(4.9)sqrt(4+3.1^2)

L=25.3077, not the right answer

Second attempt:

Diagram here.
I tried to find the tangential velocity as shown:

∠2 = arctan (3.10/2.00) = 57.17°

Vtan = 4.9cos(45) / cos(57.17) = 6.390914 m/s

ω=v/r=6.390914/Sqrt(4+3.1^2) = 1.73234 rad/s

Then I'm not really sure how to proceed after this because I'm not entirely sure what equation to use for I.

Third attempt:

Vtan=v ∴

L=1.4(6.390914)(sqrt(4+3.1^2) = 8.94728 m/s, not the right answer

Is my diagram wrong or am I missing something very important?

The magnitude of the angular momentum is L=mvr sin(theta), theta is the angle between the position vector and velocity.. Find theta. You almost have done.

The angular momentum is the vector (cross-) product of the position vector with the linear momentum. \vec L = m\vec r \times\vec v.You have the position vector 2\vec i +3.1 \vec j. You can write up the velocity vector with its components, perform the product and determine the magnitude of the vector L
ehild

 

[penguinnnnnx5]

There was a response about finding the cross product of mr x v. I did this already, but I guess I made some small mistake the first time around. I have found the correct answer. Thank you whoever you are (I forgot your username, but I remember it was something like ehlib.

 

[ehild]

It was me, but I changed my post after I saw your drawing. So all is good at last

ehild

 

[penguinnnnnx5]

The magnitude of the angular momentum is L=mvr sin(theta), theta is the angle between the position vector and velocity.. Find theta. You almost have done.

The angular momentum is the vector (cross-) product of the position vector with the linear momentum. \vec L = m\vec r \times\vec v.You have the position vector 2\vec i +3.1 \vec j. You can write up the velocity vector with its components, perform the product and determine the magnitude of the vector L



ehild

Yes, I saw that earlier... I also tried doing that the first time around; however, I must have made a minor mistake somewhere because I found the correct answer using this method. Thank you very much! Here is my work for those looking for the solution:

Position vector is:
2.00i + 3.10 j

Multiply that by m=1.4 kg, which gives:

2.89i + 4.34 j

Then you must find the x and y components of the velocity. Since it is at a 45-degree angle, both components are equal to each other:

4.9(cos 45) = 3.4648 m/s

Which gives us the velocity vector:

3.4648i + 3.4648j

You then find the cross product of mr x v which will give L = -5.335792.
Since it's asking for magnitude, the answer is 5.335792.

 

[penguinnnnnx5]

Thank you very much, sir. :)

 

[ehild]

You are welcome!
You get the same result from L=mrvsin(theta)

ehild