g A computer is reading data from a rotating CD-ROM. At a point that is 0.0189 m from the center of the disk, the centripetal acceleration i

Question

g A computer is reading data from a rotating CD-ROM. At a point that is 0.0189 m from the center of the disk, the centripetal acceleration is 241 m/s2. What is the centripetal acceleration at a point that is 0.0897 m from the center of the disc?

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Thạch Thảo 5 years 2021-07-13T09:38:35+00:00 1 Answers 129 views 0

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  1. Jezebel
    0
    2021-07-13T09:40:28+00:00
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    Answer:

    the centripetal acceleration at a point that is 0.0897 m from the center of the disc is 1143.8 m/s²

    Explanation:

    Given the data in the question;

    centripetal acceleration a_c₁ = 241 m/s²

    radius r₁ = 0.0189 m

    radius r₂ = 0.0897 m

    centripetal acceleration a_c₂ = ? m/s²

    since the rotational period will be the same for the two disk,

    we use the centripetal acceleration formula a_c = (4π²r/T²) to find the rotational period for the first disk.

    a_c₁ = (4π²r₁/T²)

    make T² subject of formula

    T² = 4π²r₁ / a_c₁

    we substitute

    T² = ( 4 × π² × 0.0189 )  / 241  

    T² = 0.00309602528 s²

    Now we use the same formula to find a_c₂

    a_c₂ = ( 4π²r₂ / T² )

    we substitute

    a_c₂ = ( 4 × π² × 0.0897 )  / 0.00309602528

    a_c₂ = 1143.8 m/s²

    Therefore, the centripetal acceleration at a point that is 0.0897 m from the center of the disc is 1143.8 m/s²

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