Respuesta :
Answer:
[tex]0.0377kgm^2/s[/tex]
Explanation:
If [tex]x[/tex] is a measured observable, then [tex]dx[/tex] is the error/uncertainty
Angular momentum,
[tex]L=\frac{1}{2}\omega MR^2= \frac{1}{2}\times21.5\times1.10\times0.250^2kgm^2/s=0.7391kgm^2/s[/tex]
To find the error in [tex]L[/tex], that is [tex]dL[/tex], we can follow the following steps:
Let us take log (base [tex]e[/tex]) both sides,
[tex]lnL=ln\omega+lnM+2lnR[/tex]
Now taking the differential of the above equation and putting in the respective values, we get,
[tex]\frac{dL}{L}=\frac{d\omega}{\omega} +\frac{dM}{M} +2\frac{dR}{R} =\frac{0.04}{21.5} +\frac{0.01}{1.10} +2\frac{0.005}{0.250}=0.051[/tex]
Therefore, uncertainty in L = [tex]0.051\times0.7391kgm^2/s=0.0377 kgm^2/s[/tex]
Her answer for angular momentum L with its uncertainty will be 0.74 ± 0.038 kgm²/s
We have been provided with following data:
M = 1.10kg and ΔM = ± 0.01kg
R = 0.250m and ΔR = ± 0.005m
ω = 21.5 rad/s and Δω = ± 0.04 rad/s
The angular momentum L is given by:
[tex]L=\frac{1}{2}I\omega[/tex]
where I=MR² is the moment of inertia and ω is the angular velocity, so
[tex]L=\frac{1}{2}MR^2\omega=0.5\times1.1\times(0.25)^2\times21.5=0.74kgm^2/s[/tex]
It is given that uncertainties are independent so taking ln of the above equation:
[tex]lnL=lnM+2lnR+ln\omega[/tex]
taking the derivative of the above equation we get:
[tex]\frac{\Delta L}{L}=\frac{\Delta M}{M}+ 2\times\frac{\Delta R}{R} +\frac{\Delta \omega}{\omega}\\ \\ =\frac{0.01}{1.1} +2\times\frac{0.005}{0.25}+ \frac{0.04}{21.5} \\\\\frac{\Delta L}{L}=0.051\\\\\Delta L=0.051\times0.74=0.038kgm^2/s[/tex]
So here measurement for L will be:
L=0.74 ± 0.038 kgm²/s
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