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?

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 a₁ = 241 m/s²

radius r₁ = 0.0189 m

radius r₂ = 0.0897 m

centripetal acceleration a₂ = ? m/s²

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

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

a₁ = (4π²r₁/T²)

make T² subject of formula

T² = 4π²r₁ / a₁

we substitute

T² = ( 4 × π² × 0.0189 )  / 241  

T² = 0.00309602528 s²

Now we use the same formula to find a₂

a₂ = ( 4π²r₂ / T² )

we substitute

a₂ = ( 4 × π² × 0.0897 )  / 0.00309602528

a₂ = 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

newton's first law of motion

photocells convert light energy to electricity