Answer:
The magnitude of the barbell's angular momentum is 38.947 kg [tex]m^{2}[/tex]/s
Explanation:
Given the mass of the barbell [tex]m_{b}[/tex] = 20 kg
mass of the rod [tex]m_{r}[/tex] = 10 kg
Length of the rod L = 1.3 m
rotation speed of the rod = 11 rpm
= 11 rpm x 2π/1 rev x 1 min/ 60 sec = 1.152 rad/sec
Angular momentum (L) = Iω............................1
I is the moment of inertia and ω is the angular sped.
As shown in the equation angular momentum is a product of angular speed and moment of inertia, the moment of inertia can be calculated thus;
I = [tex][\frac{1}{12} m_{r} L^{2} ] +2 [m_{b}(0.5L)^{2} ][/tex]
I =[ [tex]\frac{1}{12}[/tex] x 10 x[tex]1.3^{2}[/tex]] + 2[20 x(0.5 x 1.3[tex])^{2}[/tex]]
I = 1.408 + 32.4
I = 33.808 kg [tex]m^{2}[/tex]
Therefore, the moment of inertia is 33.808 and inputting the value in equation 1 we have;
L = 33.808 x 1.152
L = 38.947 kg [tex]m^{2}[/tex]/s
Therefore, the angular momentum of the barbell is 38.947 kg [tex]m^{2}[/tex]/s
