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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?

Sagot :

nuhulawal20

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[tex]_c[/tex]₁ = 241 m/s²

radius r₁ = 0.0189 m

radius r₂ = 0.0897 m

centripetal acceleration a[tex]_c[/tex]₂ = ? m/s²

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

we use the centripetal acceleration formula a[tex]_c[/tex] = (4π²r/T²) to find the rotational period for the first disk.

a[tex]_c[/tex]₁ = (4π²r₁/T²)

make T² subject of formula

T² = 4π²r₁ / a[tex]_c[/tex]₁

we substitute

T² = ( 4 × π² × 0.0189 )  / 241  

T² = 0.00309602528 s²

Now we use the same formula to find a[tex]_c[/tex]₂

a[tex]_c[/tex]₂ = ( 4π²r₂ / T² )

we substitute

a[tex]_c[/tex]₂ = ( 4 × π² × 0.0897 )  / 0.00309602528

a[tex]_c[/tex]₂ = 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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