i dont understand why we use "dM". What is dM specifically? Is it the bit of mass that we take (to calculate the moment of interia), and the reason it has the "d" at the front is because this "d" represents the rate of change? If that's the case, what is the rate of change describing? Rate of speed/velocity? thanks!
For the first case: We know Density(ρ) = Mass/Volume But for this case the ring i.e hoop has negligible thickness, similar to a straight line Thus, we consider linear mass density(ρ) =Mass/length =M/L (Length of the ring is equal to 2πr) Thus, Mass = density × length = ρ × L = ρ × 2πr Thus, for very small point on the circumference of the ring i.e hoop the mass distributed is dm and length is dr, so we can write as, dm = ρ × dr Thus I = ∫ dI I = ∫ dm R² I = R² ∫ dm I = R² ∫ ρdr I = R²ρ ∫dr Limits for integration is 0 to 2πr I = R²ρ 2πr I = R²(M/2πr)2πr I = MR²
d(put variable) represents a infinitely small value Say you have drawn a y vs x graph...and it's curved...increasing upwards. The slope must be changing at every point. But if you magnify it infinitely, you can see a straight line ( Same as drawing a circles by joining the two points of a line) This magnified portion is actually very small. Hence the 'd' comes into play. It helps us to find variable changes Here, the mass is distributed over the circumference (2πr) So, if the total mass is M Mass per unit circumference= M/2πr For very small portion like dx, the mass is small too. So it's taken to be dM. Hope it helps
Thanks for the explanation sir. I was also confused since I recently started Physics course. It is just so confusing that those integrals in physics and Riemann sum in calculus is interchangeable. @@anubhabsinha8117
very good explanation, thank you, you helped a lot!
👌Top Notch Explanation
i dont understand why we use "dM". What is dM specifically? Is it the bit of mass that we take (to calculate the moment of interia), and the reason it has the "d" at the front is because this "d" represents the rate of change? If that's the case, what is the rate of change describing? Rate of speed/velocity? thanks!
For the first case:
We know Density(ρ) = Mass/Volume
But for this case the ring i.e hoop has negligible thickness, similar to a straight line
Thus, we consider
linear mass density(ρ)
=Mass/length
=M/L
(Length of the ring is equal to 2πr)
Thus,
Mass = density × length
= ρ × L
= ρ × 2πr
Thus, for very small point on the circumference of the ring i.e hoop the mass distributed is dm and length is dr, so we can write as,
dm = ρ × dr
Thus I = ∫ dI
I = ∫ dm R²
I = R² ∫ dm
I = R² ∫ ρdr
I = R²ρ ∫dr
Limits for integration is 0 to 2πr
I = R²ρ 2πr
I = R²(M/2πr)2πr
I = MR²
d(put variable) represents a infinitely small value
Say you have drawn a y vs x graph...and it's curved...increasing upwards. The slope must be changing at every point. But if you magnify it infinitely, you can see a straight line ( Same as drawing a circles by joining the two points of a line) This magnified portion is actually very small. Hence the 'd' comes into play. It helps us to find variable changes
Here, the mass is distributed over the circumference (2πr)
So, if the total mass is M
Mass per unit circumference= M/2πr
For very small portion like dx, the mass is small too. So it's taken to be dM.
Hope it helps
Very helpful!!!😊😊😊@@techlogo365
Thanks for the explanation sir. I was also confused since I recently started Physics course. It is just so confusing that those integrals in physics and Riemann sum in calculus is interchangeable. @@anubhabsinha8117
Really amazing expalanation
Thanks
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