Two children, each with mass m = 11.0 kg, sit on opposite ends of a narrow board with length L = 4.5 m, width W = 0.30 m, and mass M = 9.4 kg. The board is pivoted at its center and is free to rotate in a horizontal circle without friction. What is the rotational inertia of the board plus the children about a vertical axis through the center of the board?
What is the magnitude of the angular momentum of the system if it is rotating with an angular speed of 2.89 rad/s?
While the system is rotating, the children pull themselves toward the center of the board until they are half as far from the center as before. What is the resulting angular speed?
What is the change in kinetic energy of the system as a result of the children changing their positions?
Moment of inertia of the board respectively the pivot point is
Io = m(L^2 + W^2)/12 = 9.4(4.5^2+0.3^2)/12 = 15.93 kg•m^2,
Moment of inertia of the system (board+2 point mass-ñhildren) respectively
the pivot point is
I = Io + 2mR^2 = 15.93 + 2•11•(4.5/2)^2 = 127.305 kg•m^2
An angular momentum
L = I•ω =127.305•2.89 = 367.9 kg•m^2/s.
In the second position: the separation of each point mass (child)
from the pivot point is r = L/4, then,
I1 = Io + 2mr^2 = 15.93+ 2•11•(4.5/4)^2 = 43.77 kg•m^2.
According to the law of conservation of
the angular momentum of the system
L = Iω = I1•ω1.
ω1= Iω/ I1 = 367.9/43.77=8.4 rad/s.
ΔKE = KE2- KE1 = I1(ω1)^2/2- I(ω)^2/2 =
= 43.77•(8.4)^2/2 - 127.305•(2.89)^2/2 =
= 1548-532=1016 J.
To find the rotational inertia of the board plus the children about a vertical axis through the center of the board, we can use the parallel axis theorem. The parallel axis theorem states that the rotational inertia about an axis parallel to an axis through the center of mass is given by the sum of the rotational inertia about the center of mass and the product of the mass and the square of the distance between the two axes.
In this case, we have two children with mass m = 11.0 kg each, and a board with mass M = 9.4 kg. The length of the board is L = 4.5 m. The rotational inertia of each child is given by I_child = (1/3) * m * W^2, where W is the width of the board (0.30 m).
The rotational inertia of the board about its center of mass is given by I_board = (1/12) * M * (L^2 + W^2).
Using the parallel axis theorem, the total rotational inertia of the board plus the children about a vertical axis through the center of the board is:
I_total = I_board + 2 * I_child
Now we can substitute the values into the formula to find the rotational inertia.
I_total = (1/12) * 9.4 * (4.5^2 + 0.3^2) + 2 * (1/3) * 11.0 * 0.3^2
Now we can calculate the value of I_total.
Next, to find the magnitude of the angular momentum of the system, we can use the formula:
L = I_total * ω
where ω is the angular speed, given as 2.89 rad/s.
Substituting the values into the formula, we can calculate the magnitude of the angular momentum.
Finally, when the children pull themselves towards the center, the moment of inertia changes. According to the conservation of angular momentum, the angular momentum of the system remains constant.
So, we can write:
I_total * ω_before = I_total * ω_after
The children are now half as far from the center, so the new distance is L/2. We can calculate the new moment of inertia as described earlier using this new distance, and the angular speed is what we need to find.
Finally, to calculate the change in kinetic energy of the system as a result of the children changing their positions, we can use the formula:
ΔKE = KE_final - KE_initial
where KE_final is the final kinetic energy of the system and KE_initial is the initial kinetic energy of the system.
The kinetic energy of a rotating system is given by:
KE = (1/2) * I_total * ω^2
Using this formula, we can calculate the initial and final kinetic energy and then find the change in kinetic energy.
By following these steps, you should be able to find the rotational inertia, magnitude of the angular momentum, resulting angular speed, and change in kinetic energy for the given system.
To find the rotational inertia of the board plus the children about a vertical axis through the center of the board, we can use the parallel-axis theorem. The rotational inertia of the board alone can be calculated as:
I_board = (1/12) * M * L^2
The rotational inertia of each child can be calculated as:
I_child = (1/12) * m * L^2
Since the children are at opposite ends of the board, we can use the parallel-axis theorem to find the rotational inertia of each child about the center of the board:
I_child' = I_child + m * (0.5L)^2
The total rotational inertia of the board plus the children is then:
I_total = 2 * I_child' + I_board
Substituting the given values:
I_total = 2 * [(1/12) * m * L^2 + m * (0.5L)^2] + (1/12) * M * L^2
Next, we can calculate the magnitude of the angular momentum (L) using the formula:
L = I_total * ω
Substituting the given angular speed (ω = 2.89 rad/s) into the formula will give us the answer.
To find the resulting angular speed after the children pull themselves towards the center of the board, we can use the principle of conservation of angular momentum. According to this principle:
I1 * ω1 = I2 * ω2
Where I1 and I2 are the initial and final rotational inertias, and ω1 and ω2 are the initial and final angular speeds.
Initially, the children are at a distance of L/2 from the center of the board. When they pull themselves toward the center, their new distance from the center becomes L/4.
Using the parallel-axis theorem, we can find the new rotational inertia:
I2 = 2 * [(1/12) * m * L^2 + m * (0.25L)^2] + (1/12) * M * L^2
Finally, we can calculate the change in kinetic energy of the system using the formula:
ΔKE = KE_final - KE_initial
The initial kinetic energy is given by:
KE_initial = 0.5 * I_total * ω1^2
The final kinetic energy is given by:
KE_final = 0.5 * I_total * ω2^2
Substituting the initial and final angular speeds into the formulas will give us the change in kinetic energy.