A merry-go-round with r = 4m and a perfect frictionless bearing is pushed with a force of 24 N by a young girl. She pushes with a constant force that is oriented tangentially to the edge of the merry-go-round. After she pushes the merry-go-round through 14 full rotations (at which point she lets go) it is spinning with an angular speed of 3 rad/s.

a. What is the moment of inertia of the merry-go-round?
b. After the girl lets go, a 20 kg boy jumps onto the merry-go-round and sticks halfway between the center of the edge. How fast are the merry-go-round and the boy spinning after he lands?

ah, it turns out I took this course in 1955 and have had lots of practice since. I do not need to do your homework. However you do. Try the problems then comeback to us with specific things you are stuck on.

The first part is moment or torque = moment of inertia * angular acceleration, alpha
with constant alpha:
omega = omega initial + alpha * t
angle = initial angle + omega initial*t + (1/2) alpha * t^2
The second part is conservation of angular momentum with the moment of inertia increased by the mass of the boy times the square of distance from center.

By the way to find the time it is easy to use the average velocity (3/2 = 1.5 radians/sec) for 14 *2 pi radians

To find the answers to the questions, we need to apply the principles of rotational dynamics. Let's break it down step by step:

a. To determine the moment of inertia of the merry-go-round, we can use the formula:

Moment of inertia (I) = (angular momentum (L)) / (angular speed (ω))

Angular momentum (L) = moment of inertia (I) × angular speed (ω)

The angular momentum is the product of the moment of inertia and the angular speed. Since the merry-go-round is initially at rest, the angular momentum is zero. After 14 full rotations, the angular speed is 3 rad/s. Therefore, we can set up the equation as follows:

0 = I × 3 rad/s

Solving for I:

I = 0

The moment of inertia for the merry-go-round is zero. This means it has either an extremely low mass or a distribution where all the mass is concentrated at the center.

b. After the girl lets go, the conservation of angular momentum is applied. The initial angular momentum of the system is equal to the final angular momentum. Since the girl is not mentioned, her angular momentum is zero.

Initial angular momentum (Linitial) = Final angular momentum (Lfinal)

The angular momentum of the system can be written as:

Linital = Lfinal

(merry-go-round moment of inertia × merry-go-round angular speed) + (boy moment of inertia × boy angular speed) = (merry-go-round moment of inertia + boy moment of inertia) × final angular speed

As mentioned, the boy sticks halfway between the center and the edge, which means he is 2 meters away from the center (r = 4 m). The moment of inertia for the merry-go-round can be calculated using the equation for a disk:

Moment of inertia (I) = (1/2) × mass × radius^2

With a radius of 4 m and a mass m, the moment of inertia for the merry-go-round is:

I_merry-go-round = (1/2) × m × (4 m)^2 = 8m

For the boy, since he sticks halfway between the center and the edge, his distance from the center is 2 m. The moment of inertia for the boy can be calculated using the equation for a point mass at a distance:

Moment of inertia (I) = mass × distance^2

With a mass of 20 kg and a distance of 2 m, the moment of inertia for the boy is:

I_boy = 20 kg × (2 m)^2 = 80 kg·m^2

The final angular speed is given as 3 rad/s, but since the question asks for both the merry-go-round and the boy's angular speeds, let's assign variables to those speeds:

Merry-go-round angular speed = ω_merry-go-round
Boy angular speed = ω_boy

Now we can set up the angular momentum equation:

(8m × ω_merry-go-round) + (80 kg·m^2 × ω_boy) = (8m + 80 kg·m^2) × 3 rad/s

Rearranging the equation to solve for both ω_merry-go-round and ω_boy:

8m × ω_merry-go-round + 80 kg·m^2 × ω_boy = 24m + 240 kg·m^2

Since we don't have a specific mass value, we can solve in terms of "m":

8 × ω_merry-go-round + 80 × ω_boy = 24 + 240

Simplifying the equation:

8 × ω_merry-go-round + 80 × ω_boy = 264

Dividing through by 8:

ω_merry-go-round + 10 × ω_boy = 33

Since we still have two variables, we need one more equation to solve the system. This equation will come from the fact that the boy sticks halfway between the center and the edge, which means his distance is half of the radius.

By using the conservation of linear momentum:

(Initial moment of inertia × initial angular speed) = (Final moment of inertia × final angular speed)

The initial moment of inertia is just the moment of inertia for the merry-go-round:

(8m × 0) = (8m + 80 kg·m^2) × 3 rad/s

Simplifying the equation:

0 = 24m + 240 kg·m^2

Rearranging:

240kg·m^2 = 24m

Dividing through by 24:

10kg·m^2 = m

Now we have a relationship between the mass of the system and the moment of inertia for the merry-go-round and the boy. Substituting this into our previous equation:

ω_merry-go-round + 10 × ω_boy = 33

ω_merry-go-round + 10 × (ω_merry-go-round/2) = 33

ω_merry-go-round + 5 × ω_merry-go-round = 33

Simplifying:

6 × ω_merry-go-round = 33

Dividing through by 6:

ω_merry-go-round = 5.5 rad/s

Finally, to find the boy's angular speed:

ω_boy = (ω_merry-go-round) / 2

ω_boy = 5.5 rad/s / 2

ω_boy = 2.75 rad/s

Therefore, after the boy lands, both the merry-go-round and the boy are spinning at an angular speed of 5.5 rad/s and 2.75 rad/s, respectively.