Calculate the rotational inertia of a meter stick, with mass 0.58 kg, about an axis perpendicular to the stick and located at the 43 cm mark. (Treat the stick as a thin rod.)
I know I = (1/12)(M)(L^2) and then I need to use the parallel-axis theorem so I need to add Mh^2 but that is not working. I get 0.156 doing this method but it is not correct.
Check your units (are they meters?) and your decimal point.
The center of the stick is at the 0.50 m mark and the moment of inertia about that point is (1/12)ML^2 = 0.0483 kg m^2.
(L = 1.0 m and M = 0.4833 kg)
Using the parallel axis theorem, the moment of inertia about an axis h = 0.07 m away is higher by
M*h^2 = 0.0028 kg/m^2, for a total of 0.4861 kg/m^2
Ah, I see what I may have done wrong. I was not using .07m for h. Unfortunately, 0.4861 for the overall answer still seems to be wrong.
Oops I found the problem, it seems you were off by a decimal point. I did the math and got .051 for a final answer, thank you for the help.
I made a typo error saying that I used a mass of M = 0.4833 kg. I actually used 0.58 kg
I may have made a computational error somewhere but I don't see where. .
(1/12)(.58)(1^2) = .0483
+
(.58)(.07^2) = .0028
So final answer about .0511 if that helps finding out your error.
You are right. I moved a decimal point and added a wrong (1/2)ML^2 value. Thanks for catching that.
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A uniform meter stick (a ruler 100 cm long) is supported by a fulcrum at its 20 cm mark balances when a 200 g?mass is suspended at 0cm.Mass of meter of stick is at the middle 50cm mark what is the mass of the meter stick in grams.
Just need to know how to set this up can some one walk show me so I know for another similiar problem I have.
To correctly calculate the rotational inertia of the meter stick using the parallel-axis theorem, you need to consider two components: the rotational inertia about the center of mass of the stick and the additional rotational inertia when considering the axis located at the 43 cm mark.
Let's break down the steps:
1. Calculate the rotational inertia about the center of mass (CM) using the formula I_cm = (1/12) * M * L^2, where M is the mass of the meter stick and L is its length.
Given:
M = 0.58 kg
L = 1 meter (100 cm)
So, I_cm = (1/12) * 0.58 kg * (100 cm)^2
Convert centimeters to meters:
I_cm = (1/12) * 0.58 kg * (1 m)^2
I_cm = (1/12) * 0.58 kg * 1 m^2
I_cm = 0.0483 kg m^2
2. Calculate the additional rotational inertia caused by the axis located at the 43 cm mark using the parallel-axis theorem. This theorem states that I_new = I_cm + M * h^2, where M is the mass of the meter stick and h is the distance between the axis of rotation and the center of mass of the stick.
Given:
M = 0.58 kg
h = 43 cm (0.43 m)
I_new = I_cm + M * h^2
I_new = 0.0483 kg m^2 + 0.58 kg * (0.43 m)^2
I_new = 0.0483 kg m^2 + 0.58 kg * 0.1849 m^2
I_new = 0.0483 kg m^2 + 0.106982 kg m^2
I_new = 0.155282 kg m^2
So, the correct rotational inertia of the meter stick about the axis perpendicular to the stick and located at the 43 cm mark is approximately 0.1553 kg m^2.