A CD player rotates at a variable speed so that a laser can scan pits and lands on the disk’s bottom surface at a constant tange
1 answer:
Answer:
W = 0.384 J
Explanation:
Work and energy in the rotational movement are related
W = ΔK = - K₀
W = ½ I ² - 1 /2 I w₀²
where W isthe work, I is the moment of inertia and w angular velocity
With the system part of the rest the initial angular speed is zero (w₀ = 0)
The angular and linear quantities are related
v = w r
w = v / r
Let's replace
W = ½ I (v / r)²
Let's calculate
W = ½ 1.2 10⁻⁴ 1.2² / (1.5 10⁻²)²
W = 0.384 J
W = 38.4 J
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Answer:
w = 4,786 rad / s , f = 0.76176 Hz
Explanation:
For this problem let's use the concept of angular momentum
L = I w
The system is formed by the two discs, during the impact the system remains isolated, we have the forces are internal, this implies that the external torque is zero and the angular momentum is conserved
Initial Before sticking
L₀ = 0 + I₂ w₂
Final after coupling
= (I₁ + I₂) w
The moments of inertia of a disk with an axis of rotation in its center are
I = ½ M R²
How the moment is preserved
L₀ =
I₂ w₂ = (I₁ + I₂) w
w = w₂ I₂ / (I₁ + I₂)
Let's reduce the units to the SI System
d₁ = 60 cm = 0.60 m
d₂ = 40 cm = 0.40 m
f₂ = 200 min-1 (1 min / 60 s) = 3.33 Hz
Angular velocity and frequency are related.
w₂ = 2 π f₂
w₂ = 2π 3.33
w₂ = 20.94 rad / s
Let's replace
w = w₂ (½ M₂ R₂²) / (½ M₁ R₁² + ½ M₂ R₂²)
w = w₂ M₂ R₂² / (M₁ R₁² + M₂ R₂²)
Let's calculate
w = 20.94 8 0.40² / (12 0.60² + 8 0.40²)
w = 20.94 1.28 / 5.6
w = 4,786 rad / s
Angular velocity and frequency are related.
w = 2π f
f = w / 2π
f = 4.786 / 2π
f = 0.76176 Hz