1. The problem statement, all variables and given/known data
A 2.0-m measuring stick of mass 0.175 kg is resting on a table. A mass of 0.500 kg is attached to the stick at a distance of 74.0 cm from the center. Both the stick and the table surface are frictionless. The stick rotates with an angular speed of 5.30 rad/s.
(a) If the stick is pivoted about an axis perpendicular to the table and passing through its center, what is the angular momentum of the system?
(b) If the stick is pivoted about an axis perpendicular to it and at the end that is furthest from the attached mass, and it rotates with the same angular speed as before, what is the angular momentum of the system?
2. Relevant equations
L=Iω
I=MR^2
I=(1/12)ML^2
I=(1/3)ML^2
3. The attempt at a solution
Okay so for the first question I used the moment of inertia formula and treated the mass as a point mass.
(0.500*0.740^2)=0.27
Plus the moment of inertia for the stick(rod) since it passing through the center I used I=(1/12)ML^2
(1/12)(0.175*2^2)=0.058
Then I added and multiplied by the angular velocity
0.27+0.058=0.33*5.3=1.7 kg*m/s
Now this answer was right however when I did the second one I repeated the same process but used I= (1/3)ML^2 because it at the end of the rod.
So I did
(0.500*0.740^2)=0.27
Then
(1/3)(0.175*2^2)=0.23
Then I added and multiplied by the angular velocity
0.27+0.23=0.50*5.3=2.65 rounded to 2.7 kg*m/s
It is saying both 2.65 and 2.7 were wrong. Where did I go wrong?
To find the angular momentum of the system, we need to calculate the moment of inertia (I) for each scenario and multiply it by the angular velocity (ω). Let's go through the steps again for both scenarios:
(a) When the stick is pivoted about an axis perpendicular to the table and passing through its center:
1. Calculate the moment of inertia of the attached mass using the formula I = MR²:
I_attached = (0.500 kg) * (0.740 m)² = 0.2586 kg·m²
2. Calculate the moment of inertia of the stick (rod) using the formula I = (1/12)ML²:
I_stick = (1/12) * (0.175 kg) * (2.0 m)² = 0.0583 kg·m²
3. Add the two moment of inertia values:
I_total = I_attached + I_stick = 0.2586 kg·m² + 0.0583 kg·m² = 0.3169 kg·m²
4. Multiply the total moment of inertia by the angular velocity:
L = I_total * ω = 0.3169 kg·m² * 5.30 rad/s = 1.6797 kg·m²/s (approx. 1.68 kg·m²/s)
Therefore, the angular momentum of the system in scenario (a) is approximately 1.68 kg·m²/s.
(b) When the stick is pivoted about an axis perpendicular to it and at the end that is furthest from the attached mass:
1. Calculate the moment of inertia of the attached mass using the formula I = MR²:
I_attached = (0.500 kg) * (0.740 m)² = 0.2586 kg·m²
2. Calculate the moment of inertia of the stick (rod) using the formula I = (1/3)ML²:
I_stick = (1/3) * (0.175 kg) * (2.0 m)² = 0.23 kg·m²
3. Add the two moment of inertia values:
I_total = I_attached + I_stick = 0.2586 kg·m² + 0.23 kg·m² = 0.4886 kg·m²
4. Multiply the total moment of inertia by the angular velocity:
L = I_total * ω = 0.4886 kg·m² * 5.30 rad/s = 2.59018 kg·m²/s (approx. 2.59 kg·m²/s)
Therefore, the angular momentum of the system in scenario (b) is approximately 2.59 kg·m²/s.
It seems that your calculations for scenario (b) were correct. The discrepancy in the answer might be due to a rounding error during the calculations or if the system conserves angular momentum, the answer should be the same for both scenarios.