Convert g*cm^2 to kg*m^2: Explanation & Help

AI Thread Summary
To convert a moment of inertia from g*cm^2 to kg*m^2, first recognize that the conversion involves both mass and length squared. The conversion factors are 0.001 kg/g and 0.01 m/cm, which must be squared for the area unit, resulting in a factor of 0.000001 (or 10^-6) when applied to g*cm^2. Therefore, the correct approach is to multiply the initial value by 0.000001 to obtain the equivalent in kg*m^2. This method ensures that units cancel appropriately, allowing for accurate conversion. Understanding these squared conversion factors is crucial for successful unit transformations in physics problems.
azure.hubris
Messages
9
Reaction score
0
Hello,
I need to convert a moment of intertia value given in g*cm^2 to kg*m^2, and was hoping someone could give me a run down of the method behind these kind of conversions. Do I simply assume that since g to kg is a factor 0.001 and cm to m is 0.01, that I can apply a factor of 0.00001 to my initial value? I feel that that method would be fine for say, g*cm to kg*cm, but I'm not sure how to account for the squared dimension. Any help/explanation would be appreciated. Thanks.
 
Physics news on Phys.org
In general when converting units set your quantity to be converted and your conversion factor up as a multiplication of fractions. For example from m/s to cm/s
10 \frac m s * 100 \frac {cm} {m} = 10*100 \frac m s * \frac {cm} {m} = 1000 \frac {cm} s
Note that in this example the m in the numerator cancels the m in the denominator, leaving only cm in the numerator.
 
Last edited:
okay, i see what you mean, I've used that technique before, but the confusion came from two places for me.
1) that I'm dealing with a squared quantity, so would the conversion be 1000cm^2/m^2?
2) that the it's g*cm^2, rather than g/cm^2, leaving me wondering how to actually set up the calculation.

i'm sure it's quite obvious how to do this, but the way i set it up, the units don't cancel out. i need to be certain of this conversion before i can begin to do the problem, since it's for an online assignment for which i have only one attempt. could you possibly lay out how you'd set it up for the case of my calculation? i know it's not usually acceptable to 'give' the solution away in this forum, but the physics of the problem are not a concern, just this one calculation. okay, thanks again for your help.
 
The method is the same just square the conversion factors. So

10m^2 = 10 m^2 * (100 \frac {cm} m)^2 = 10 * 10^4 m^2 \frac {cm^2} {m^2} = 10^5 cm^2
 
thanks for your help!
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Trying to understand the logic behind adding vectors with an angle between them'
My initial calculation was to subtract V1 from V2 to show that from the perspective of the second aircraft the first one is -300km/h. So i checked with ChatGPT and it said I cant just subtract them because I have an angle between them. So I dont understand the reasoning of it. Like why should a velocity be dependent on an angle? I was thinking about how it would look like if the planes where parallel to each other, and then how it look like if one is turning away and I dont see it. Since...
Back
Top