A child, hunting for his favorite wooden horse, is running on the ground around the edge of a stationary merry-go-round. The angular speed of the child has a constant value of 0.264 rad/s. At the instant the child spots the horse, one-quarter of a turn away, the merry-go-round begins to move (in the direction the child is running) with a constant angular acceleration of 0.00500 rad/s2. What is the shortest time it takes for the child to catch up with the horse?
To find the shortest time it takes for the child to catch up with the horse, we need to analyze the motion of both the child and the merry-go-round.
Let's start by finding the angular position of the child at the instant he spots the horse. We know that the child's angular speed is 0.264 rad/s, and the time it takes for the child to run one-quarter of a turn is given by:
θ = ω*t
0.25*2*pi = 0.264*t
Solving for t, we find:
t = (0.25*2*pi) / 0.264
t ≈ 2.38 s
So, it takes approximately 2.38 seconds for the child to spot the horse.
Next, let's consider the angular motion of the merry-go-round. It starts with an initial angular velocity of 0 rad/s and has an angular acceleration of 0.00500 rad/s^2. We want to find out how long it takes for the merry-go-round to match the child's angular position.
Using the equation of angular motion:
θ = ω0*t + (1/2) * α * t^2
We know that the initial angular velocity (ω0) is 0, the angular acceleration (α) is 0.00500 rad/s^2, and the final position (θ) is 0.25*2*pi. We need to solve for t:
0.25*2*pi = 0.5 * 0.00500 * t^2
Simplifying the equation:
t^2 = (0.25*2*pi) / (0.5*0.00500)
t^2 = 2*pi / (0.00500)
t^2 ≈ 1256
Taking the square root of both sides:
t ≈ √(1256)
t ≈ 35.4 s
Therefore, it takes approximately 35.4 seconds for the merry-go-round to catch up with the child.
To find the shortest time it takes for the child to catch up with the horse, we need to add the time it takes for the child to spot the horse (2.38 s) to the time it takes for the merry-go-round to catch up with the child (35.4 s):
Shortest time = 2.38 s + 35.4 s ≈ 37.8 s
So the shortest time it takes for the child to catch up with the horse is approximately 37.8 seconds.
To solve this problem, we need to analyze the motion of both the child and the merry-go-round.
Let's denote:
- θ_child = the angular position of the child relative to the stationary merry-go-round
- θ_horse = the angular position of the horse relative to the stationary merry-go-round
- ω_child = the initial angular speed of the child (0.264 rad/s)
- α_merry = the angular acceleration of the merry-go-round (0.00500 rad/s^2)
At the instant the child spots the horse, the angular position of the child is one-quarter of a turn away from the horse. Since one complete turn is 2π radians, one-quarter of a turn is (1/4) * 2π = π/2 radians. So we have θ_child = π/2.
We want to find the time it takes for the child to catch up with the horse, which is the time it takes for θ_child to equal θ_horse.
Since the merry-go-round starts moving when the child spots the horse, let's denote t as the time that has elapsed since the merry-go-round started moving.
The angular position of the horse is given by θ_horse = ω_horse * t, where ω_horse is the angular speed of the horse. We need to find ω_horse.
Since the child is running in the same direction as the merry-go-round, the relative speed between the child and the horse is the difference of their angular speeds. Therefore, the angular speed of the horse relative to the stationary merry-go-round is given by:
ω_horse = ω_child + ω_merry,
where ω_merry is the angular speed of the merry-go-round.
Since the merry-go-round has a constant angular acceleration, we can relate its angular speed to time using the kinematic equation:
ω_merry = α_merry * t.
Now we can substitute ω_merry into the previous equation:
ω_horse = ω_child + α_merry * t.
In order for the child to catch up with the horse, we need θ_child to equal θ_horse. So we can write:
θ_child = θ_horse,
π/2 = ω_horse * t.
Now we can substitute ω_horse into the equation:
π/2 = (ω_child + α_merry * t) * t.
This is a quadratic equation in t, so we can solve it for t.
First, let's rearrange the equation:
0 = α_merry * t^2 + ω_child * t - π/2.
Now, we can use the quadratic formula to solve for t:
t = (-ω_child ± √(ω_child^2 + 4 * α_merry * (π/2))) / (2 * α_merry).
Plugging in the values:
t = (-0.264 ± √(0.264^2 + 4 * 0.00500 * (π/2))) / (2 * 0.00500).
Evaluating this equation, we get two solutions for t:
t ≈ -2.78 s and t ≈ 5.34 s.
Since time cannot be negative, we take the positive value.
Therefore, the shortest time it takes for the child to catch up with the horse is approximately 5.34 seconds.
The child reaches the horse when the angular positions of the child and horse, measured from initial angular position of the child, are the same.
0.264 t = (pi/2) + (1/2)(0.005)t^2
Solve for t in seconds.
The pi/2 term represents the quarter turn lead of the horse.