A computer is reading data from a rotating CD-ROM. At a point that is 0.0314m from the center of the disk, the centripetal acceleration is 165m/s2. What is the centripetal acceleration at a point that is 0.0805m from the center of the disc?
The Correct Answer and Explanation is:
To find the centripetal acceleration at a point farther from the center of the CD-ROM, we must understand how centripetal acceleration relates to the radius.
🔷 Given:
- Radius 1, r1=0.0314 mr_1 = 0.0314 \, \text{m}
- Centripetal acceleration at r1r_1, ac1=165 m/s2a_{c1} = 165 \, \text{m/s}^2
- Radius 2, r2=0.0805 mr_2 = 0.0805 \, \text{m}
- Need to find: ac2=?a_{c2} = ?
🔷 Concept:
Centripetal acceleration aca_c is given by: ac=ω2ra_c = \omega^2 r
Where:
- ω\omega is the angular velocity in radians per second (rad/s)
- rr is the radius
If the CD-ROM spins at constant angular velocity (which is typically the case for rotating disks), then ω\omega is the same at all points on the disk. This means centripetal acceleration is directly proportional to radius: ac2ac1=r2r1\frac{a_{c2}}{a_{c1}} = \frac{r_2}{r_1}
🔷 Plug in values:
ac2165=0.08050.0314\frac{a_{c2}}{165} = \frac{0.0805}{0.0314} ac2=165×0.08050.0314a_{c2} = 165 \times \frac{0.0805}{0.0314} ac2=165×2.5637≈422.01 m/s2a_{c2} = 165 \times 2.5637 \approx 422.01 \, \text{m/s}^2
✅ Final Answer:
422 m/s2\boxed{422 \, \text{m/s}^2}
🔷 Explanation (300 words):
Centripetal acceleration occurs whenever an object moves in a circular path. It is the inward acceleration required to keep the object moving in that circle. The formula ac=ω2ra_c = \omega^2 r shows that this acceleration depends both on the angular speed (ω\omega) of the object and its distance from the center of the circle (rr).
In this problem, the CD-ROM spins at a fixed rate, meaning every point on the disk completes a rotation in the same amount of time. Because of that, the angular velocity ω\omega is constant throughout the disk. When angular velocity is constant, centripetal acceleration increases linearly with the radius—double the distance from the center, and the centripetal acceleration doubles.
Here, the point at 0.0314 m from the center experiences 165 m/s² of centripetal acceleration. To find the acceleration at 0.0805 m, we used a ratio: since ac∝ra_c \propto r, we multiplied 165 by the ratio 0.08050.0314\frac{0.0805}{0.0314}, which is approximately 2.5637. This gives a new acceleration of approximately 422 m/s².
This relationship is particularly important in devices like CD-ROMs, where data is read at varying distances from the center. Engineers must account for changes in acceleration and velocity across the disc to ensure smooth operation.