A 25 kg child stands 2.5 m from the center of a frictionless merry‐go‐round, which has a 200 kg*m^2 moment of inertia and is spi
1 answer:
Answer:
Explanation:
a ) Time period T = 2 s
Angular velocity ω = 2π / T
= 2π / 2 = 3.14 rad /s
Initial moment of inertia I₁ = 200 + mr²
= 200 + 25 x 2.5²
=356.25
Final moment of inertia
I₂ = 200 + 25 X 1.5 X 1.5
= 256.25
b ) We apply law of conservation of momentum
I₁ X ω₁ = I₂ X ω₂
ω₂ = I₁ X ω₁ / I₂
Putting the values
ω₂ = 4.365 rad s⁻¹
c ) Increase in rotational kinetic energy
=1/2 I₂ X ω₂² - 1/2 I₁ X ω₁²
.5 X 256.25 X 4.365² - .5 X 356.25 X 3.14²
= 684.95 J
This energy comes from work done against the centripetal pseudo -force.
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Answer:
v = 19.33 m / s South
Explanation:
To solve this exercise we must use the conservation of momentum, for which we must define a system formed by the two cars, therefore the forces during the collision are internal and therefore the moment is conserved.
Since it is a vector quantity, we are going to work on each axis, the x axis is in the East-West direction
initial instant. Before the crash
p₀ = m 0 + M v₂ₓ
final instant. Right after the crash
p_f = (m + M) vₓ
p₀ = 0_pf
M v₂ₓ = (m + M) vₓ
In this case m is the mass of the car and M the mass of the SUV
vₓ = v₂ₓ (1)
in the Y axis (North - South direction)
initial instant
p₀ = m v_{1y} + M 0
final moment
p_f = (m + M) v_y
p₀ = p_f
m v_{1y} + M 0 = (m + M) v_y
v_y = (2)
With these speeds we can use the relationship between work and the variation of kinetic energy, in this part the two cars are already united.
W = ΔK
friction force work is
W = - fr d
the friction force is described by the equation
fr = μ N
Newton's second law
N-W = 0
N = W
we substitute
W = - μ (m + M) g d
as the car stops the final kinetic energy is zero and
the initial kinetic energy is
K₀ = ½ (m + M) v²
we substitute
- μ (m + M) g d = 0 - ½ (m + M) v²
μ g d = ½ v²
v² = 2 μ g d
the distance traveled can be found with the Pythagorean theorem
d =
d =
d = 8.40 m
let's calculate the speed
v² = 2 0.75 9.8 8.40
v = √123.48
v = 11.11 m / s
this velocity is in the direction of motion so we can use trigonometry to find the angles
tan θ = y / x
θ = tan⁻¹ y / x
θ = tan⁻¹ (-5.48 / -6.37)
θ = 40.7º
Since the two magnitudes are negative, this angle is in the third quadrant, measured from the positive side of the x-axis in a counterclockwise direction.
θ'= 180 + 40.7
θ’= 220.7º
In the exercise they indicate the the sedan moves in the y-axis, therefore
sin θ'= v_y / v
v_y = v sin 220.7
v_y = 11.11 sin 220.7
v_y = -7.25 m / s
the negative sign indicates that it is moving south
To find the speed we substitute in equation 2
v_y =
v_{1y} = v_ y
let's calculate
v_{1y} = -7.25
v_{1y} = - 19.33 m/s
therefore the speed of the sedan is v = 19.33 m / s with a direction towards the South