Answer:
a. [tex]I=2.77x10^{-8} kg*m^2[/tex]
b. [tex]K=4.37 x10^{-6} N*m[/tex]
Explanation:
The inertia can be find using
a.
[tex]I = m*r^2[/tex]
[tex]m = 0.95 g * \frac{1 kg}{1000g}=9.5x10^{-4} kg[/tex]
[tex]r=0.54 cm * \frac{1m}{100cm} =5.4x10^{-3}m[/tex]
[tex]I = 9.5x10^{-4}kg*(5.4x10^{-3}m)^2[/tex]
[tex]I=2.77x10^{-8} kg*m^2[/tex]
now to find the torsion constant can use knowing the period of the balance
b.
T=0.5 s
[tex]T=2\pi *\sqrt{\frac{I}{K}}[/tex]
Solve to K'
[tex]K = \frac{4\pi^2* I}{T^2}=\frac{4\pi^2*2.7702 kg*m^2}{(0.5s)^2}[/tex]
[tex]K=4.37 x10^{-6} N*m[/tex]
