Rotational Motion: Average angular acceleration of a CD

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The discussion focuses on calculating the average angular acceleration of a compact disc (CD) during its 74-minute playback time. The initial and final angular velocities are provided, with values of 50 rad/s and 21.6 rad/s, respectively. The average angular acceleration is calculated using the formula α = (ωf - ωi) / Δt, leading to a result of -0.006 rad/s². The user initially submitted an answer with insufficient significant figures, which was corrected to meet the three significant figure requirement. The final resolution confirms that proper formatting of significant figures is crucial for accurate submissions.
adangerousdriver
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Homework Statement


"A compact disc (CD) stores music in a coded pattern of tiny pits 10−7m deep. The pits are arranged in a track that spirals outward toward the rim of the disc; the inner and outer radii of this spiral are 25.0 mm and 58.0 mm, respectively. As the disc spins inside a CD player, the track is scanned at a constant linear speed of 1.25 m/s."
"What is the average angular acceleration of a maximum-duration CD during its 74.0-min playing time? Take the direction of rotation of the disc to be positive."

rinside=.025m
routside=.058m
v=1.25m/s
Calculated from previous parts of this problem:
ωinside=50rad/s
ωoutside=21.6rad/s
t=74min=74*60s

Homework Equations


α=(ωfi)/(Δt)

The Attempt at a Solution


I figured average angular acceleration would be calculated by subtracting the initial angular velocity from the final and dividing by the time interval, so I plugged in and performed this calculation:

(21.6rad/s-50rad/s)/(74*60s-0s)

to get -0.006 rad/s2. This is online homework and it told me my answer is incorrect. I can't tell what I'm doing wrong. I've tried submitting both -0.006 and 0.006.
 
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Your given values have 3 significant figures. What you've submitted only has one.
 
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gneill said:
Your given values have 3 significant figures. What you've submitted only has one.
Welp that fixed it. Thank you!
 
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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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