Determine experimentally which rotational axis yields the maximum rotational inertia (i.e., moment of inertia) and which yields
1 answer:
Answer:
the maximum is I₁ axis of rotation at the end
the minimum moment is I₂ axis of rotation at the center of mass
Explanation:
For this exercise we use the definition moment of inertia
I = ∫ r² dm
for bodies of high symmetry it is tabulated; In this case we can approximate a broomstick to a thin rod, the moment of inertia with respect to a perpendicular axis when varying are
at one end
I₁ = ⅓ mL²
in in center
I₂ = m L²
There is another possible axis of rotation around the axis of the broom, in this case we have a solid cylinder
I₃ = m r²
remember that the diameter of the broom is much smaller than its length, therefore this moment of inertia is very small
when examining the different moments of inertia:
the maximum is I₁ axis of rotation at the end
the minimum moment is I₂ axis of rotation at the center of mass
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The H field is in units of amps/meter. It is sometimes called the auxiliary field. It describes the strength (or intensity) of a magnetic field. The B field is the magnetic flux density. It tells us how dense the field is. If you think about a magnetic field as a collection of magnetic field lines, the B field tells us how closely they are spaced together. These lines (flux linkages) are measured in a unit called a Weber (Wb). This is the analog to the electric charge, the Coulomb. Just like electric flux density (the D field, given by D=εE) is Coulombs/m², The B field is given by Wb/m², or Tesla. The B field is defined to be μH, in a similar way the D field is defined. Thus B is material dependent. If you expose a piece of iron (large μ) to an H field, the magnetic moments (atoms) inside will align in the field and amplify it. This is why we use iron cores in electromagnets and transformers.
So if you need to measure how much flux goes through a loop, you need the flux density times the area of the loop Φ=BA. The units work out like
Φ=[Wb/m²][m²]=[Wb], which is really just the amount of flux. The H field alone can't tell you this because without μ, we don't know the "number of field" lines that were caused in the material (even in vacuum) by that H field. And the flux cares about the number of lines, not the field intensity.
I'm way into magnetic fields, my PhD research is in this area so I could go on forever. I have included a picture that also shows M, the magnetization of a material along with H and B. M is like the polarization vector, P, of dielectric materials. If you need more info let me know but I'll leave you alone for now!
So if you need to measure how much flux goes through a loop, you need the flux density times the area of the loop Φ=BA. The units work out like
Φ=[Wb/m²][m²]=[Wb], which is really just the amount of flux. The H field alone can't tell you this because without μ, we don't know the "number of field" lines that were caused in the material (even in vacuum) by that H field. And the flux cares about the number of lines, not the field intensity.
I'm way into magnetic fields, my PhD research is in this area so I could go on forever. I have included a picture that also shows M, the magnetization of a material along with H and B. M is like the polarization vector, P, of dielectric materials. If you need more info let me know but I'll leave you alone for now!
Refer to the diagram shown below.
In order for the balloon to strike the professor's head, th balloon should drop by 18 - 1.7 = 16.3 m in the time at the professor takes to walk 1 m.
The time for the professor to walk 1 m is
t = (1 m)/(0.45 m/s) = 2.2222 s
The initial vertical velocity of the balloon is zero.
The vertical drop of the balloon in 2.2222 s is
h = (1/2)*(9.8 m/s²)*(2.2222 s)² = 24.197 m
Because 24.97 > 16.3, the balloon lands in front of the professor, and does not hit the professor.
The time for the balloon to hit the ground is
(1/2)*(9.8)*t² = 18
t = 1.9166 s
The time difference is 2.2222 - 1.9166 = 0.3056 s
Within this time interval, the professor travels 0.45*0.3056 = 0.175 m
Therefore the balloon falls 0.175 m in front of the professor.
Answer:
The balloon misses the professor, and falls 0.175 m in front of the professor.
In order for the balloon to strike the professor's head, th balloon should drop by 18 - 1.7 = 16.3 m in the time at the professor takes to walk 1 m.
The time for the professor to walk 1 m is
t = (1 m)/(0.45 m/s) = 2.2222 s
The initial vertical velocity of the balloon is zero.
The vertical drop of the balloon in 2.2222 s is
h = (1/2)*(9.8 m/s²)*(2.2222 s)² = 24.197 m
Because 24.97 > 16.3, the balloon lands in front of the professor, and does not hit the professor.
The time for the balloon to hit the ground is
(1/2)*(9.8)*t² = 18
t = 1.9166 s
The time difference is 2.2222 - 1.9166 = 0.3056 s
Within this time interval, the professor travels 0.45*0.3056 = 0.175 m
Therefore the balloon falls 0.175 m in front of the professor.
Answer:
The balloon misses the professor, and falls 0.175 m in front of the professor.