A merry-go-round rotates at the rate of.2rev/s with an 80kg man standing at a point 2m from the axis ofrotation. (A)What is the new angular speed when the man walksto a point 1m from the center? Assume that the merry-go-round is a solid 25kg cylinder of radius of 2m.
B. calculate the change in kinetic energy due to man moving
C. how do you account for this change in kinetic energy?
angular momentum (L) = Inertia (I) * angular velocity (w)
Using conservation of angular momentum (L):
Inertia initial total (Ii) * angular velocity initial (wi) = Inertia final total (If) * angular velocity final (wf)
Inertia of disks (such as the merry-go-round) = .5 * mass (m) * radius squared (r)
Inertia of a point-mass (such as the person) = m * r^2
with this you should get the following:
Li=Lf
Ii*wi=lf*wf
(I + Ii)*wi=(I + If)*wf
(.5MR^2 + mri^2)wi=(.5MR^2 + mrf^2)wf
where
M=mass of the merry-go-round
R=radius of the merry-go-round
m=mass of the person
ri=initial radius of the person
re-arrange and solve for wf for part a
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change in KE(rotational) is KEf-KEi
KE(rot) = .5*I*w^2
you should get a positive change. This makes sense because if you think of radius between the person and the pivot-point as the distance associated with Potential Energy, as the distance decreases, the speed along with Kinetic energy should increase due to conservation of energy.
PEi + KE(rot)i = PEf + KE(rot)f
big + small = small + big
change in KE = big - small = positive change
I hope that makes sense :)
A. To find the new angular speed when the man walks to a point 1m from the center, we can use the principle of conservation of angular momentum.
The initial angular momentum of the merry-go-round with the man is given by:
L1 = I1 ω1 (equation 1)
where L1 represents the initial angular momentum, I1 is the moment of inertia of the system (merry-go-round + man), and ω1 is the initial angular speed.
The final angular momentum of the system with the man closer to the center is given by:
L2 = I2 ω2 (equation 2)
where L2 represents the final angular momentum, I2 is the moment of inertia of the system (with the man closer to the center), and ω2 is the final angular speed.
According to the conservation of angular momentum, L1 = L2.
Since the merry-go-round is modeled as a solid cylinder, the moment of inertia can be calculated as:
I = 1/2 * m * r^2
Substituting the values into equations 1 and 2:
I1 * ω1 = I2 * ω2
(1/2 * m1 * r1^2) * ω1 = (1/2 * m2 * r2^2) * ω2
where m1 = mass of the man = 80 kg
m2 = mass of the merry-go-round = 25 kg
r1 = initial distance of the man from the center = 2 m
r2 = final distance of the man from the center = 1 m
Simplifying the equation, we get:
(40 * 2^2) * 0.2 = (25 * 1^2) * ω2
160 * 0.2 = 25 * ω2
32 = 25 * ω2
ω2 = 32/25 = 1.28 rad/s
Therefore, the new angular speed when the man walks to a point 1m from the center is approximately 1.28 rad/s.
B. To calculate the change in kinetic energy due to the man moving, we can use the formula:
Change in Kinetic Energy = Final Kinetic Energy - Initial Kinetic Energy
The initial kinetic energy of the system is given by:
KE1 = 1/2 * I1 * ω1^2
The final kinetic energy of the system is given by:
KE2 = 1/2 * I2 * ω2^2
Substituting the values into the equations:
KE1 = 1/2 * (1/2 * m1 * r1^2) * ω1^2
KE2 = 1/2 * (1/2 * m2 * r2^2) * ω2^2
Change in Kinetic Energy = KE2 - KE1
Simplifying the equation, we get:
Change in Kinetic Energy = 1/2 * [(1/2 * m2 * r2^2 * ω2^2) - (1/2 * m1 * r1^2 * ω1^2)]
Plugging in the given values:
Change in Kinetic Energy = 1/2 * [(1/2 * 25 * 1^2 * 1.28^2) - (1/2 * 80 * 2^2 * 0.2^2)]
Calculating this equation would give us the change in kinetic energy.
C. The change in kinetic energy is accounted for by the work done by the external forces acting on the system. As the man moves closer to the center, the moment of inertia of the system decreases, causing a decrease in its kinetic energy. This decrease in kinetic energy is balanced by the work done by the forces acting on the system, such as friction or the man's own effort in moving towards the center.
A) To find the new angular speed when the man walks to a point 1m from the center, we can use the law of conservation of angular momentum. Angular momentum is defined as the product of moment of inertia and angular velocity.
The initial angular momentum of the system can be represented as:
L₁ = I₁ω₁
where L₁ is the initial angular momentum, I₁ is the initial moment of inertia, and ω₁ is the initial angular velocity.
The final angular momentum of the system can be represented as:
L₂ = I₂ω₂
where L₂ is the final angular momentum, I₂ is the final moment of inertia, and ω₂ is the final angular velocity.
According to the conservation of angular momentum, the initial and final angular momenta should be equal:
L₁ = L₂
Now, let's calculate the initial and final moment of inertia for the system.
For the initial moment of inertia, we can use the formula for a solid cylinder:
I₁ = (1/2)mr₁²
where m is the mass of the merry-go-round (25 kg in this case) and r₁ is the initial distance of the man from the axis of rotation (2 m).
Plugging in the values, we have:
I₁ = (1/2)(25 kg)(2 m)²
I₁ = 100 kg·m²
For the final moment of inertia, we will use the same formula, but with the final distance of the man from the axis of rotation (1 m):
I₂ = (1/2)mr₂²
where r₂ is the final distance of the man from the axis of rotation (1 m).
Plugging in the values, we have:
I₂ = (1/2)(25 kg)(1 m)²
I₂ = 12.5 kg·m²
Now, we can equate the initial and final angular momenta:
I₁ω₁ = I₂ω₂
Solving for ω₂, we get:
ω₂ = (I₁/I₂)ω₁
Plugging in the values, we have:
ω₂ = (100 kg·m²) / (12.5 kg·m²) * 0.2 rev/s
ω₂ = 1.6 rev/s
Therefore, the new angular speed when the man walks to a point 1m from the center is 1.6 rev/s.
B) To calculate the change in kinetic energy due to the man moving, we need to find the initial and final kinetic energies of the system.
The initial kinetic energy of the system can be represented as:
K₁ = (1/2)I₁ω₁²
where K₁ is the initial kinetic energy.
Plugging in the values, we have:
K₁ = (1/2)(100 kg·m²)(0.2 rev/s)²
K₁ = 2 J
The final kinetic energy of the system can be represented as:
K₂ = (1/2)I₂ω₂²
where K₂ is the final kinetic energy.
Plugging in the values, we have:
K₂ = (1/2)(12.5 kg·m²)(1.6 rev/s)²
K₂ = 19.2 J
The change in kinetic energy can be calculated as the difference between the final kinetic energy and the initial kinetic energy:
ΔK = K₂ - K₁
ΔK = 19.2 J - 2 J
ΔK = 17.2 J
Therefore, the change in kinetic energy due to the man moving is 17.2 J.
C) The change in kinetic energy is accounted for by the work done on the system. When the man moves closer to the center, he exerts a force to push himself inward, doing work on the system against the rotational inertia. This work is converted into a change in kinetic energy, resulting in the increase in rotational speed. In other words, the man's movement transfers energy to the merry-go-round, causing it to rotate faster and therefore increasing its kinetic energy.