A merry-go-round starts from rest and accelerates uniformly over 29.0 s to a final angular velocity of 5.35 rev/min.
(a) Find the maximum linear speed of a person sitting on the merry-go-round 4.50 m from the center.
(b) Find the person's maximum radial acceleration.
(c) Find the angular acceleration of the merry-go-round.
(d) Find the person's tangential acceleration.
C=6.28*4.5 = 28.3m/rev.=Circumference.
a. V=5.35rev/min * 28.3m/rev=151m/min.
c. a = (Vf - Vo) / t,
a = (5.35 - 0)rev/min / (29/60)min =
11.1m/min^2.
To find the maximum linear speed of a person sitting on the merry-go-round, we can use the formula:
v = r * ω
Where:
v = linear speed
r = radius
ω = angular velocity
For this problem, the radius (r) is given as 4.50 m and the angular velocity (ω) is given as 5.35 rev/min. To convert the angular velocity to radians per second, we use the conversion factor: 1 rev = 2π radians.
So the angular velocity in rad/s is:
ω = 5.35 rev/min * (2π radians/rev) * (1 min/60 s) = 0.556 rad/s
Now we can plug the values into the formula to find the linear speed:
v = (4.50 m) * (0.556 rad/s) = 2.50 m/s
Therefore, the maximum linear speed of a person sitting 4.50 m from the center of the merry-go-round is 2.50 m/s.
To find the person's maximum radial acceleration, we can use the formula:
a_r = r * ω^2
Where:
a_r = radial acceleration
r = radius
ω = angular velocity
Using the given values, we have:
a_r = (4.50 m) * (0.556 rad/s)^2 = 1.48 m/s^2
Therefore, the person's maximum radial acceleration is 1.48 m/s^2.
To find the angular acceleration of the merry-go-round, we can use the formula:
α = Δω / Δt
Where:
α = angular acceleration
Δω = change in angular velocity
Δt = change in time
The angular velocity changes from 0 (starting from rest) to 5.35 rev/min in 29.0 s. To convert the final angular velocity to rad/s, we use the conversion factor mentioned earlier.
So the change in angular velocity is:
Δω = 5.35 rev/min * (2π radians/rev) * (1 min/60 s) = 0.556 rad/s
Now we can plug the values into the formula to find the angular acceleration:
α = (0.556 rad/s) / (29.0 s) = 0.0192 rad/s^2
Therefore, the angular acceleration of the merry-go-round is 0.0192 rad/s^2.
To find the person's tangential acceleration, we can use the formula:
a_t = r * α
Where:
a_t = tangential acceleration
r = radius
α = angular acceleration
Using the given values, we have:
a_t = (4.50 m) * (0.0192 rad/s^2) = 0.0864 m/s^2
Therefore, the person's tangential acceleration is 0.0864 m/s^2.