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an old 78 rpm record rotates through an angle of 320° as it slows down uniformly from 78.0 rpm to 22.8 rpm. what is the magnitude of the angular acceleration of the record?

User Rod Dewell
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1 Answer

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Final answer:

To find the angular acceleration of a record slowing down from 78 rpm to 22.8 rpm through an angle of 320°, we use the kinematic equation for rotational motion after converting rpm to rad/s and degrees to radians, resulting in the angular acceleration expressed in rad/s².

Step-by-step explanation:

We need to find the angular acceleration of a record that is slowing down uniformly from 78 rpm to 22.8 rpm over an angle of 320°. First, let's convert the angular velocities from rpm to rad/s:

  • ω₀ (initial angular velocity) = 78 rpm × (2π rad/1 rev) × (1 min/60 s) = 78 × 2π / 60 rad/s
  • ω (final angular velocity) = 22.8 rpm × (2π rad/1 rev) × (1 min/60 s) = 22.8 × 2π / 60 rad/s

Next, we'll use the following kinematic equation for rotational motion:

ω² = ω₀² + 2αΘ

Where ω is the final angular velocity, ω₀ is the initial angular velocity, α is the angular acceleration, and Ψ is the angle the record has turned through in radians. We need to convert the angle from degrees to radians:

Θ = 320° × (π rad / 180°)

By rearranging the formula to solve for α, we get:

α = (ω² - ω₀²) / (2Θ)

After plugging in the values and performing the calculation, we will have the magnitude of the angular acceleration, expressed in rad/s².

User Holman
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