A 3-kg rock swings in a circle of radius 5 m. making one complete trip every 2 seconds. What is the centripetal acceleration of the rock?
- oubaasLv 77 months ago
ω = 2PI/T = 2PI/2 = PI rad/sec
ac = ω^2*r = 3.1416^2*5 = 49.35 m/sec^2
- electron1Lv 77 months ago
Centripetal acceleration = v^2 ÷ r
During the two seconds, the rock moves a distance of the circumference of a circle that has a radius.5 meters..
d = 2 * π * 5 = 10 * π meters
v = d ÷ t = 10 * π ÷ 2 = 5 * π
The speed of the rock is approximately 15.7 m/s.
Centripetal acceleration = 25 * π^2 ÷ 5 = 5 * π^2
This is approximately 49.3 m/s^2. I hope this is helpful for you.
- 7 months ago
Centripetal acceleration is v²/r where v is velocity given by time taken to traverse the circumference in one rotation which is 2π*5 / 2 m/s = 5π² m/s.
So we substitute (5π)² /5 to get Centripetal acceleration is 5π² m/s²
Given that the mass of the rock is 3 kg, we may determine the force due to centripetal acceleration using the formula Force = mass x acceleration
F = ma, = 3*5π² = 148.0 m/s²
- Some BodyLv 77 months ago
The radius is 5 m, so the circumference is:
C = 2πr
C = 10π
The rock completes one revolution in 2 seconds, so its velocity is:
v = C / t
v = 10π / 2
v = 5π
We can now find the centripetal acceleration:
a = v² / r
a = (5π)² / 5
a = 5π²
a ≈ 49.3 m/s²