Given Information:
Diameter of the cylindrical tub = d = 50 cm = 0.50 m
Acceleration = α = 3g
Required Information:
1. Rotation rate in rev/min = ω = ?
2. Tangential speed in m/s = v = ?
Answer:
1. ω = 103.5 rev/min
2. v = 2.71 m/s
Explanation:
We know that centripetal acceleration is given by
α = ω²r
Where ω is the angular speed or rotation rate and r is the radius.
The relation between diameter and radius is given by
r = D/2
r = 0.50/2
r = 0.25 m
Since it is given that the acceleration is equal to 3g where g is the gravitational acceleration 9.81 m/s².
α = ω²r
3g = ω²r
ω² = 3g/r
ω = √(3g/r)
ω = √(3*9.81/0.25)
ω = 10.84 rad/s
To convert rad/s into rev/s divide it by 2π
ω = 10.84/2π
ω = 1.752 rev/s
To convert rev/s into rev/min multiple it by 60
ω = 1.752*60
ω = 103.5 rev/min
Therefore, the rotation rate is 103.5 rev/min
2. The tangential speed can be found using
v = ωr
Where ω is the rotation rate in rad/s and r is the radius.
v = 10.84*0.25
v = 2.71 m/s
Therefore, the tangential speed is 2.71 m/s
