A person of mass 55 kg stands at the center of a rotating merry-go-round platform of radius 3.4 m and moment of inertia 670 kg·m2. The platform rotates without friction with angular velocity 1.0 rad/s. The person walks radially to the edge of the platform.
(a) Calculate the angular velocity when the person reaches the edge.
(b) Calculate the rotational kinetic energy of the system of platform plus person before the person's walk.
(c) Calculate the rotational kinetic energy of the system of platform plus person after the person's walk.
Please Help I have no clue where to even start! :(
Don't worry, I can help you with that! Let's break down each part of the question step by step.
(a) To calculate the angular velocity when the person reaches the edge, we can use the principle of conservation of angular momentum. Angular momentum is a conserved quantity in the absence of external torques. Initially, the person is at the center of the platform, and when they walk to the edge, their moment of inertia changes.
The initial moment of inertia (I1) of the system (platform plus person) is the moment of inertia of the platform alone, given as 670 kg·m^2.
The final moment of inertia (I2) of the system is the moment of inertia of the platform plus the person. To calculate it, we can use the parallel axis theorem, which states that the moment of inertia about an axis parallel and a distance "d" away from a known axis is equal to the moment of inertia about the known axis plus the mass multiplied by the square of the distance.
In this case, the final moment of inertia (I2) would be equal to the moment of inertia of the platform plus the moment of inertia of the person. The moment of inertia of a point mass (person) rotating around a fixed axis at a distance "r" is given by the formula I = mr^2, where "m" is the mass of the person and "r" is the distance from the rotation axis.
So, I2 = I1 + mr^2.
Given that the person's mass is 55 kg and the radius of the platform is 3.4 m, we have:
I2 = 670 kg·m^2 + (55 kg)(3.4 m)^2.
Now, we can apply the principle of conservation of angular momentum. The initial angular momentum (L1) of the system is equal to the final angular momentum (L2) after the person walks to the edge.
Angular momentum is given by the formula L = Iω, where "ω" is the angular velocity.
L1 = I1ω1, and L2 = I2ω2.
Since angular momentum is conserved, we can equate L1 and L2:
I1ω1 = I2ω2.
Plugging in the values we have, we get:
670 kg·m^2 * 1.0 rad/s = (670 kg·m^2 + (55 kg)(3.4 m)^2) * ω2.
Solving for ω2 will give us the angular velocity when the person reaches the edge.
(b) The initial rotational kinetic energy of the system can be calculated using the formula:
KE1 = (1/2) * I1 * ω1^2.
Plugging in the given values, we can find the answer.
(c) The final rotational kinetic energy of the system after the person's walk can be calculated using the formula:
KE2 = (1/2) * I2 * ω2^2.
Once we have obtained ω2 from part (a), we can substitute it into this equation along with the values we have to calculate the answer.
So, by following these steps, you'll be able to find the answers to parts (a), (b), and (c) of the question. Good luck!
This is a conservation of angular momentum problem.
Now, the formula for angular momentum is:
L = Iw
So basically, L before = L after:
I1w1 = I2w2
The trick to solving these is to figure out what the change in moment of inertia is, and then apply the concept of conservation of angular momentum to it.
With a person on a merry-go-round, the moment of inertia would be the moment of inertia of the person plus the moment of inertia of the merry-go-round.
In this case the person starts at the center of the rotating merry-go-round, and so I think we are to say their moment of inertia is zero or negligible (it would be small), but then they move to the outside edge of the merry-go-round, and in the after picture, have a significant moment of inertia. The moment of inertia of the merry-go-round is the same before and after and given as 670 kgm2