A merry-go-round starts from rest and accelerates uniformly over 26.0 s to a final angular velocity of 5.50 rev/min.
(a) Find the maximum linear speed of a person sitting on the merry-go-round 4.25 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.
An angular velocity is
ω =2πn= 5.5•2π/60= 0.58 rad/s.
Linear speed
v= ωR=0.58•4.25=2.45 m/s,
Radial acceleration
a = ω^2•R=(0.58)^2•4.25=1.43 m/s^2
Since ω = εt , an angular acceleration
ε = ω/t=0.58/26=2.23•10^-2 rad/s^2,
Tangential acceleration
a = ε•R= 2.23•10^-2•4.25=0.095 m/s^2
(a) To find the maximum linear speed of a person sitting on the merry-go-round 4.25 m from the center, we can use the formula:
Linear speed (v) = radius (r) × angular velocity (ω)
Given:
Radius (r) = 4.25 m
Angular velocity (ω) = 5.50 rev/min
First, we need to convert the angular velocity from rev/min to rad/s. Since 1 revolution is equal to 2π radians, and 1 minute is equal to 60 seconds, we can calculate:
ω = (5.50 rev/min) × (2π rad/1 rev) × (1 min/60 s)
ω = 5.50 × 2π / 60 rad/s
Now, we can calculate the linear speed:
v = (4.25 m) × (5.50 × 2π / 60 rad/s)
v = (4.25 m) × (5.50 × 2π / 60 rad/s)
v ≈ 3.84 m/s
Therefore, the maximum linear speed of a person sitting on the merry-go-round 4.25 m from the center is approximately 3.84 m/s.
(b) To find the person's maximum radial acceleration, we can use the formula:
Radial acceleration (ar) = r × angular acceleration (α)
We are not given the angular acceleration (α) directly, so let's find it using the formula for uniform acceleration:
Angular acceleration (α) = change in angular velocity (Δω) / time (t)
Given:
Final angular velocity (ωf) = 5.50 rev/min
Time (t) = 26.0 s
First, let's convert the final angular velocity to radians per second:
ωf = 5.50 rev/min × 2π rad/1 rev × 1 min/60 s
ωf = 5.50 × 2π / 60 rad/s
Now, let's calculate the change in angular velocity (Δω):
Δω = ωf - ωi
Since the merry-go-round starts from rest, the initial angular velocity (ωi) is 0.
Δω = 5.50 × 2π / 60 rad/s - 0 rad/s
Δω = 5.50 × 2π / 60 rad/s
Now, we can calculate the angular acceleration:
α = Δω / t = (5.50 × 2π / 60 rad/s) / 26.0 s
α ≈ 0.567 rad/s^2
Finally, we can calculate the person's maximum radial acceleration:
ar = r × α = (4.25 m) × (0.567 rad/s^2)
ar ≈ 2.41 m/s^2
Therefore, the person's maximum radial acceleration is approximately 2.41 m/s^2.
(c) To find the angular acceleration of the merry-go-round, we already calculated it in part (b):
Angular acceleration (α) ≈ 0.567 rad/s^2
Therefore, the angular acceleration of the merry-go-round is approximately 0.567 rad/s^2.
(d) To find the person's tangential acceleration, we can use the formula:
Tangential acceleration (at) = r × angular acceleration (α)
Given:
Radius (r) = 4.25 m
Angular acceleration (α) ≈ 0.567 rad/s^2
We can calculate the tangential acceleration:
at = (4.25 m) × (0.567 rad/s^2)
at ≈ 2.41 m/s^2
Therefore, the person's tangential acceleration is approximately 2.41 m/s^2.
To solve this problem, we will use the equations of rotational motion. These equations relate the linear motion of an object to its angular motion.
First, let's convert the final angular velocity from rev/min to rad/s. Since 1 revolution is equal to 2π radians, we have:
Final angular velocity (ω) = 5.50 rev/min
Convert to rad/s: ω = (5.50 rev/min) * (2π rad/rev) * (1 min/60 s) = 0.575 rad/s
(a) 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 is the linear speed, r is the radius, and ω is the angular velocity.
Given: r = 4.25 m, ω = 0.575 rad/s
Substituting these values into the formula:
v = (4.25 m) * (0.575 rad/s) = 2.44 m/s
So, the maximum linear speed of the person is 2.44 m/s.
(b) The radial acceleration (ar) of the person can be calculated using the formula:
ar = r * ω^2
Given: r = 4.25 m, ω = 0.575 rad/s
Substituting these values into the formula:
ar = (4.25 m) * (0.575 rad/s)^2 = 1.14 m/s^2
Therefore, the person's maximum radial acceleration is 1.14 m/s^2.
(c) The angular acceleration (α) of the merry-go-round can be found using the equation:
ω = ω₀ + α * t
where ω₀ is the initial angular velocity, α is the angular acceleration, and t is the time.
Given: ω₀ = 0 (since the merry-go-round starts from rest), ω = 0.575 rad/s, t = 26.0 s
Rearranging the equation, we get:
α = (ω - ω₀) / t = (0.575 rad/s - 0) / 26.0 s = 0.0221 rad/s^2
Thus, the angular acceleration of the merry-go-round is 0.0221 rad/s^2.
(d) The tangential acceleration (at) of the person can be calculated using the formula:
at = r * α
Given: r = 4.25 m, α = 0.0221 rad/s^2
Substituting these values into the formula:
at = (4.25 m) * (0.0221 rad/s^2) = 0.094 m/s^2
Therefore, the person's tangential acceleration is 0.094 m/s^2.