A playground merry-go-round has a disk-shaped platform that rotates with negligible friction about a vertical axis. The disk has a mass of 210 kg and a radius of 1.8 m. A 35- kg child rides at the center of the merry-go-round while a playmate sets it turning at 0.10rpm. If the child then walks along a radius to the outer edge of the disk, how fast will the disk be turning (in rpm - but enter no unit)?
(.5*210*1.8^2)(.10)=(.5*210*1.8^2)+(35*1.8^2)(final rpm)
34.02=147.42*final rpm
34.02/147.42=final rpm
=.2308 rpm
To solve this problem, we can apply the principle of conservation of angular momentum.
The angular momentum (L) of an object is given by the product of its moment of inertia (I) and its angular velocity (ω). Mathematically, it can be expressed as:
L = I * ω
In this case, the initial angular momentum of the system is equal to the final angular momentum of the system.
The initial angular momentum (Li) of the system (disk + child) is given by the product of the moment of inertia of the disk and the initial angular velocity of the disk (ωi).
Li = I * ωi
The final angular momentum (Lf) of the system is given by the product of the moment of inertia of the disk (considering the child moved to the outer edge of the disk) and the final angular velocity of the disk (ωf).
Lf = I' * ωf
Since angular momentum is conserved, Li = Lf, we can write:
I * ωi = I' * ωf
The moment of inertia (I) of a disk is given by the formula:
I = (1/2) * m * r^2
Where m is the mass of the disk and r is the radius of the disk.
Given:
Mass of the disk (m) = 210 kg
Radius of the disk (r) = 1.8 m
Mass of the child (m_child) = 35 kg
Initial angular velocity (ωi) = 0.10 rpm
Before the child moves, the moment of inertia of the system (I) is:
I = (1/2) * (m_disk + m_child) * r^2
I = (1/2) * (210 kg + 35 kg) * (1.8 m)^2
The final moment of inertia (I') of the system after the child moves can be calculated as follows:
I' = (1/2) * m_disk * r'^2
Where r' is the new radius of the system when the child moves to the outer edge of the disk.
The radius from the center to the outer edge of the disk is equal to the sum of the radius of the disk and the distance the child walks (r_child):
r' = r + r_child
We need to calculate r_child.
To find the distance the child walks (r_child), we can use the fact that the moment of inertia remains constant. We can write:
I * ωi = I' * ωf
Simplifying, we have:
(1/2) * (210 kg + 35 kg) * (1.8 m)^2 * (0.10 rpm) = (1/2) * 210 kg * ((1.8 m + r_child)^2) * ωf
Now we can solve for r_child.
Simplifying further:
(1/2) * (245 kg) * (1.8 m)^2 * (0.10 rpm) = (1/2) * (210 kg) * ((1.8 m + r_child)^2) * ωf
Calculating, we find:
70.65 kg * m^2 * rpm = 44.1 kg * (1.8 m + r_child)^2 * ωf
Cross multiplying:
70.65 kg * m^2 * rpm = 44.1 kg * (1.8 m + r_child)^2 * ωf
Dividing both sides by 44.1 kg * (1.8 m + r_child)^2:
(70.65 kg * m^2 * rpm) / (44.1 kg * (1.8 m + r_child)^2) = ωf
Calculating this value, we find:
ωf ≈ 0.2687
Finally, converting ωf to rpm, we have:
ωf ≈ 0.2687 * (60 sec/1 min) = 16.12 rpm
Therefore, the disk will be turning at approximately 16.12 rpm when the child walks to the outer edge of the disk.
To find the final angular velocity of the disk after the child walks to the outer edge, we can use the principle of conservation of angular momentum.
The formula for angular momentum is given by L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
Since the child is at the center of the merry-go-round initially, the moment of inertia of the system is given by the moment of inertia of the disk alone, which is equal to 1/2 * mass * radius^2.
So, the initial angular momentum of the system is given by L1 = I1 * ω1, where I1 is the initial moment of inertia of the disk and ω1 is the initial angular velocity of the disk with the child at the center.
When the child walks to the outer edge of the disk, the moment of inertia of the system changes. Now, the moment of inertia is given by the moment of inertia of the disk alone, plus the moment of inertia of the child, which is approximated as a point mass (mass * distance^2).
The final angular velocity of the disk, with the child at the outer edge, is given by ω2.
According to the conservation of angular momentum, L1 = L2. So, I1 * ω1 = I2 * ω2.
Substituting the moment of inertia formulas, and solving for ω2, we get:
(1/2 * mass * radius^2) * ω1 = (1/2 * mass * radius^2 + mass * distance^2) * ω2.
Given the values:
mass of the disk (m) = 210 kg
radius of the disk (r) = 1.8 m
mass of the child (M) = 35 kg
We need to find the distance (d) the child walks along the radius from the center to the outer edge.
Using the conservation of linear momentum, we can relate the final linear velocity (V2) of the child at the outer edge with the initial linear velocity (V1) at the center:
m * V1 = (m + M) * V2.
Rearranging the equation, we get:
V2 = (m * V1) / (m + M).
Given the initial angular velocity (ω1) of the disk is 0.10 rpm, we can convert it to rad/s by multiplying it by 2π/60.
Plugging in the values, we can solve for the distance (d) using the equation:
d = V2 * Δt,
where Δt is the time it takes for the child to walk along the radius.
Finally, we can substitute the distance (d) into the formula for ω2 and solve for the final angular velocity of the disk.
Note: The specific numerical answer will depend on the calculated value of the distance (d) and the time it takes for the child to walk along the radius.