You want to do a physics experiment, so you put a box at the edge of a merry-go-round, 8.0 m in diameter and 1.5 m above the ground. The operator turns on the ride. When the merry-go-round gets to 0.34 rev/s, the box goes flying off, landing on the ground. The merry-go-round has completed 8.0 turns when the box goes flying. Assume constant angular acceleration of the merry-go-round. (a) What is the total acceleration of the box, just before flying off the merry-go-round? (b) How long is the box in the air (from the moment it leaves the merry-go-round, to the moment it hits the ground)? solve using circular motion, kinematic
To solve this problem, we will use the principles of circular motion and kinematics.
(a) Let's start by finding the angular acceleration of the merry-go-round. Angular acceleration (α) is the rate at which the angular velocity (ω) changes with time (t). In this case, the angular acceleration is constant.
Given:
Diameter of the merry-go-round (d) = 8.0 m
Radius of the merry-go-round (r) = d/2 = 8.0 / 2 = 4.0 m
Angular velocity (ω) when the box goes flying = 0.34 rev/s
To find the angular acceleration (α), we can use the equation:
ω^2 = ω0^2 + 2αθ
Where:
ω^2 is the final angular velocity squared
ω0^2 is the initial angular velocity squared
α is the angular acceleration
θ is the angle covered in radians
Since the box has completed 8.0 turns, the angle covered (θ) can be calculated as:
θ = (number of turns) x 2π
= 8.0 x 2π
= 16π radians
Substituting the given values into the equation, we have:
(0.34 rev/s)^2 = 0 + 2α(16π)
0.1156 rev^2/s^2 = 32πα
From this equation, we can determine the value of α.
(b) Now let's find the time (t) it takes for the box to hit the ground after leaving the merry-go-round. We can use the following kinematic equation:
y = y0 + v0yt - (1/2)gt^2
Where:
y is the final height of the box above the ground (which is 0 in this case)
y0 is the initial height of the box above the ground (1.5 m)
v0y is the initial vertical velocity of the box
g is the acceleration due to gravity (9.8 m/s^2)
t is the time it takes for the box to hit the ground
We need to find v0y and t. To find the initial vertical velocity (v0y), we can use the relationship between linear velocity (v) and angular velocity (ω) in circular motion:
v = rω
Substituting the given values, we have:
v0y = rω
= (4.0 m)(0.34 rev/s)
= 1.36 m/s
Now we can solve for t using the kinematic equation:
0 = 1.5 m + (1.36 m/s)t - (1/2)(9.8 m/s^2)t^2
Simplifying and rearranging the equation, we have:
4.9t^2 - 1.36t - 1.5 = 0
We can solve this quadratic equation to find the value of t using the quadratic formula or factoring.
By solving these equations, we can find both the total acceleration of the box just before flying off the merry-go-round and the time for which the box is in the air.