A merry-go-round is stationary. A dog is running on the ground just outside its circumference, moving with a constant angular speed of 0.705 rad/s. The dog does not change his pace when he sees what he has been looking for: a bone resting on the merry-go-round one third of a revolution in front of him (which means the dog has to move for additional 2/3Pi angular displacement before reaching the bone.) At the instant the dog sees the bone (t = 0), the merry-go-round begins to move in the direction the dog is running, with a constant angular acceleration of 0.0135 rad/s2. At what time will the dog reach the bone?
Hint: use rotational kinematics equations for both, dog and the merry-go-round and set them equal to each other - this means the dog has reached the bone. Then solve it for time, t.
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To find the time at which the dog reaches the bone on the merry-go-round, we need to use rotational kinematics equations for both the dog and the merry-go-round. Let's break down the steps to solve this problem:
Step 1: Find the angular displacement of the dog before reaching the bone.
Given that the dog needs to move for an additional 2/3π angular displacement before reaching the bone, we can calculate it as follows:
Angular displacement = (2/3)π radians
Step 2: Find the angular acceleration and initial angular velocity of the merry-go-round.
The merry-go-round has a constant angular acceleration of 0.0135 rad/s^2.
The initial angular velocity of the merry-go-round can be assumed to be 0 since it starts from rest.
Step 3: Find the angular displacement and initial angular velocity of the dog.
The dog has a constant angular velocity of 0.705 rad/s.
The initial angular displacement of the dog is 0 since it starts from the reference point.
Step 4: Set up the rotational kinematics equation for both the dog and the merry-go-round.
For the dog:
θ = ω*t + (1/2)*α*t^2
For the merry-go-round:
θ = ω*t + (1/2)*α*t^2
Step 5: Set the two equations equal to each other.
Since the dog reaches the bone when the angular displacement for both the dog and the merry-go-round is the same, we can equate the two equations.
θ (dog) = θ (merry-go-round)
Step 6: Substitute the values into the equation and solve for time, t.
(2/3)π = (0.705)t + (1/2)*(0.0135)t^2
This equation is a quadratic equation in terms of t. You can simplify it by multiplying through by 2 to get rid of the fraction and rearrange it to the form ax^2 + bx + c = 0.
Step 7: Solve the quadratic equation for t.
You can solve the quadratic equation using the quadratic formula or factoring, and then determine the positive value of t since a negative time doesn't make sense in this context.
Step 8: Once you have the value of t, you will have the time at which the dog reaches the bone on the merry-go-round.
By following these steps and solving the quadratic equation, you can find the time at which the dog reaches the bone on the merry-go-round.