The teacher shared a demonstration with the same students. She begin by standing on a desk, holding a book and an open piece of
paper in the air, away from her body.. She asked her students: "Which will hit the ground first if I drop them?" Based on the previous data, the students answered that they would hit the ground at the same time. When she dropped them, the book hit the ground first. Next she crumpled up the paper and repeated the experiment.
Elaborate on the two paper drops. Was there different results? Why?
A) No. In both cases, the book dropped faster than the paper because it had so much more mass.
B) Yes. The crumpled paper dropped at the same rate as the book; the flat sheet did not due to air resistance.
C) In both cases the paper fell at the same rate because in both cases the air resistance and the masses were the same.
D) Yes there was a difference in the rate at which the papers fell. The crumpled paper had the same air resistance as the book so they fell at the same rate.
Elaborate on the two paper drops. Was there different results? Why?
A) No. In both cases, the book dropped faster than the paper because it had so much more mass.
B) Yes. The crumpled paper dropped at the same rate as the book; the flat sheet did not due to air resistance.
C) In both cases the paper fell at the same rate because in both cases the air resistance and the masses were the same.
D) Yes there was a difference in the rate at which the papers fell. The crumpled paper had the same air resistance as the book so they fell at the same rate.
1 answer:
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Answer:
a) K = 0.63 J, b) h = 0.153 m
Explanation:
a) In this exercise we have a physical pendulum since the rod is a material object, the angular velocity is
w² =
where d is the distance from the pivot point to the center of mass and I is the moment of inertia.
The rod is a homogeneous body so its center of mass is at the geometric center of the rod.
d = L / 2
the moment of inertia of the rod is the moment of a rod supported at one end
I = ⅓ m L²
we substitute
w =
w =
w =
w = 4.427 rad / s
an oscillatory system is described by the expression
θ = θ₀ cos (wt + Φ)
the angular velocity is
w = dθ /dt
w = - θ₀ w sin (wt + Ф)
In this exercise, the kinetic energy is requested in the lowest position, in this position the energy is maximum. For this expression to be maximum, the sine function must be equal to ±1
In the exercise it is indicated that at the lowest point the angular velocity is
w = 4.0 rad / s
the kinetic energy is
K = ½ I w²
K = ½ (⅓ m L²) w²
K = 1/6 m L² w²
K = 1/6 0.42 0.75² 4.0²
K = 0.63 J
b) for this part let's use conservation of energy
starting point. Lowest point
Em₀ = K = ½ I w²
final point. Highest point
Em_f = U = m g h
energy is conserved
Em₀ = Em_f
½ I w² = m g h
½ (⅓ m L²) w² = m g h
h = 1/6 L² w² / g
h = 1/6 0.75² 4.0² / 9.8
h = 0.153 m