A thin rod of length 2.35 m and mass 13.9 kg is rotated at an angular speed of 3.75 rad/s around an axis perpendicular to the rod and through its center of mass. Find the magnitude of the rod's angular momentum.
I used the formula ML^2/3 * w and got 95.96 but it tells me it's wrong? I wanted a second opinion.
(1 / 12) m L^2 * omega
rod is spinning around CENTER, not end so 1/12 not 1/3
divide whatever you did by 4
Well, it seems like you've got the right idea, but maybe you made a small mistake in your calculation. Let's double-check your working!
The formula for angular momentum is L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular speed.
The moment of inertia of a thin rod rotating about an axis perpendicular to it and through its center of mass is given by I = (1/12) * m * L^2, where m is the mass of the rod and L is its length.
Plugging in the given values, we have:
m = 13.9 kg
L = 2.35 m
So, the moment of inertia is:
I = (1/12) * 13.9 kg * (2.35 m)^2
Now, substitute this moment of inertia and the given angular speed ω = 3.75 rad/s into the formula for angular momentum:
L = I * ω
Now, go ahead and calculate it accurately, and let's see if we get the same result!
Based on your provided information, the formula for the magnitude of angular momentum (L) of a rotating rod is given by:
L = Iω
Where:
I is the moment of inertia of the rod about its axis of rotation
ω is the angular speed of the rod
To calculate the moment of inertia, you need to use the formula for a thin rod rotating about an axis perpendicular to the rod and passing through its center of mass:
I = (1/12) * m * L^2
Where:
m is the mass of the rod
L is the length of the rod
Substituting the given values:
m = 13.9 kg
L = 2.35 m
ω = 3.75 rad/s
Calculating the moment of inertia:
I = (1/12) * 13.9 kg * (2.35 m)^2 = 0.823 kg·m² (rounded to three decimal places)
Now we can calculate the magnitude of the angular momentum:
L = I * ω = 0.823 kg·m² * 3.75 rad/s = 3.079 kg·m²/s (rounded to three decimal places)
Therefore, the magnitude of the rod's angular momentum is approximately 3.079 kg·m²/s.
To find the magnitude of the rod's angular momentum, you can use the formula L = I * ω, where L is the angular momentum, I is the moment of inertia, and ω is the angular speed.
The moment of inertia for a thin rod rotating about an axis perpendicular to the rod and through its center of mass is given by the formula I = (1/12) * m * L^2, where m is the mass of the rod and L is the length of the rod.
Substituting the given values:
m = 13.9 kg
L = 2.35 m
First, calculate the moment of inertia:
I = (1/12) * m * L^2
= (1/12) * 13.9 kg * (2.35 m)^2
= (1/12) * 13.9 kg * 5.5225 m^2
≈ 0.92125 kg * m^2
Next, substitute the moment of inertia and the given angular speed into the formula for angular momentum:
L = I * ω
≈ 0.92125 kg * m^2 * 3.75 rad/s
≈ 3.4547 kg * m^2/s
So, the magnitude of the rod's angular momentum is approximately 3.4547 kg * m^2/s.
Comparing this to your calculated value of 95.96 kg * m^2/s, it seems there might have been an error in your calculation. Double-check your calculations and make sure you are using the correct formulas.