Rotational Inertia and Motion problems

[PhysicsPenguin]

Hello everyone, I am trying to solve some homework but I am not entirely sure what formulas to use.

Problem) There are 3 kids that weigh 25kg each sitting in the center of a merry go round (Disk). The merry-go-round weighs 400 kg and has a diameter of 3 m. It is initially spinning at 50 rpm.

A) What is the initial rotation?

I tried using the formula 1/2 mr^2. .5(400kg)(1.5)^2 = 450kg*m/s

B) What is the initial kinetic energy of the merry go round with the kids on it?
I tried 1/2 IW^2
1/2 (450)(5.2359878)^2 which equals 6168.5 N

W would be the 50 rpm converted to radians/second which would be 5.23rad/s and I was = 450 from earlier.

C) If all the kids move to the edge of the merry go round, what will be the new angular velocity in rad/s?

I got a little lost on this, would the proper formula be L = iw. 50 = 450w and w be the answer?

D) What is the magnitude of the linear velocity of each child as they sit on the edge and it spins at the rate round in C?

V = rω
V = 1.5(Answer from C in radians/sec)

E) The kids now want to stop. they drag their feet so they each exert a force F that is tangent to the disk in a direction that causes the merry-go-round to stop. How much net work must be done by the 2 kids in order to bring the disk with them on the edge to a stop.?

Not sure how to do this one.

F) If the disk stops after 6 seconds, what is the deceleration of the disk?
Wouldn't this ω = ωo + (alpha)t
5.23(rad/s) = 0 + (alpha)(6)
5.23/6 = alpha?

G) What is the torque on the disk?
Not sure how to solve this one

H) Magnitude of the average Force F each kid needs to exert?
Not sure again how to solve this

I) How many revolutions will the disk need to turn before coming to a stop?

 

[BiGyElLoWhAt]

A) I don't know what you mean by initial rotation
B) $\frac{1}{2}I\omega ^2$ is the correct formula, and the kids are at the center (i.e. the axis of rotation, so they don't contribute to the moment of inertia) so you're good there.
C)Hint, something in this situation is conserved, and something changes when the children move outwards.
D)Right equation, but wrong numbers. See C (HA!)
E) Your looking for work, so energy, sooo...?
F) No, you have to find a different omega. Again, See C.
G) $T= R\ (cross)\ F$
H) Use what you get from F
J) Use the rotational analogue for your displacement kinematic equation

Hope this helps

 

[PhysicsPenguin]

It did help some, Thank you.

I got part A/B/C/D done I believe. I did A as finding I, (I = 1/2 mr^2).

For A I got 450 kg* m^2 /s
B: 6168.5 J
C:4.40925
D: 6.613875For G, I am not sure what you mean. The only three formulas I have that involve T are T = 1/f (frequency)
T = la and T = Delta L / Delta time.

 

[BiGyElLoWhAt]

T as in torque, not as in Period.

Torque is equal to radius cross the force.

 

[HalfLostAndSearching]

I would suggest using the formulas for rotational inertia and angular velocity to solve these problems. For problem A, the formula for rotational inertia of a disk is correct, but the units should be in kg*m^2. So the initial rotational inertia of the merry-go-round would be 1/2(400kg)(1.5m)^2 = 450kg*m^2.

For problem B, the formula for kinetic energy of a rotating object is 1/2Iω^2, where I is the moment of inertia and ω is the angular velocity in radians/second. So the initial kinetic energy of the merry-go-round with the kids on it would be 1/2(450kg*m^2)(5.23rad/s)^2 = 6168.5J.

Moving on to problem C, the formula for angular momentum is L = Iω, where I is the moment of inertia and ω is the angular velocity in radians/second. So the new angular velocity can be found by setting the initial angular momentum equal to the final angular momentum (since angular momentum is conserved in a closed system). So the calculation would be: 50rpm = (450kg*m^2)(ω) and solving for ω gives us an angular velocity of 5.23rad/s.

For problem D, the formula for linear velocity is v = rω, where r is the distance from the center of rotation (in this case, the radius of the merry-go-round) and ω is the angular velocity in radians/second. So the linear velocity of each child sitting on the edge would be 1.5m(5.23rad/s) = 7.85m/s.

Moving on to problem E, we can use the formula for work, W = Fd, where F is the force applied and d is the distance over which the force is applied. Since the kids are exerting a force tangent to the disk, the distance over which the force is applied would be the circumference of the disk, which is 2πr = 2π(1.5m) = 9.42m. So the net work done by the kids would be W = F(9.42m).

For problem F, we can use the formula for angular acceleration, α = (ωf - ωi)/t, where ωf is the final