Test Questions i got wrong, help please

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In summary, the first problem involves calculating the rotational kinetic energy of a system consisting of two masses attached to a uniform rod, while the second problem involves finding the location of the center of mass for a non-uniform density cylinder. The solution for the first problem involves using conservation of angular momentum and neglecting vertical velocities, while the solution for the second problem can be found using symmetry and treating the cylinder as a tapered rod with the same mass distribution.
  • #1
ic3man
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2 probs: Rotational Kinetic Energy & Center of Mass

Got these questions wrong on a previous test. Have a final coming up soon and i would like to be able to understand these questions:


Homework Statement



1. Two masses, m1 = 2Kg and m2 = 4 Kg are thrown horizontally with the same speed of v = 10 m/s to strike and get attached at the two ends of a uniform rod of length L = 10m, pivoted at its center of mass. The rod is capable of rotation in the horizontal plane. The masses strike the rod at the same time. The mass of the rod is M = 2 Kg. Calculate the rotational kinetic energy of system after the collision.

2. A thin cylinder of radius R and length L has a non-uniform density given by p = p0 (x/L) but uniform along the radial direction. Calculate the location of the center of mass Xcm

(p0 in the previous problem is P null, which i assume is a constant)


2. The attempt at a solution

1. I didnt even know where to begin with this one

2. If this was a uniform rod, not a cylinder i could get the answer, but i wasnt sure if i had to take into consideration that it was a cylinder. Typically in questions in my homework, it doesn't provide a rod that has R, so i was wondering if the fact that it was a cylinder was not important, althought the teacher did provide the formula for the volume of a cylinder on the test.


Thanks for all the help. Couldnt figure out these questions on a previous test and i don't desire getting similar ones wrong on the final.
 
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  • #2
For #1 the final motion will be governed by conservation of angular momentum. A lot of mechanical energy will be lost in the process. You need to find the angular momentum of each ball about the pivot point. Any vertical momentum will be overcome by the vertical forces holding the rod in place, so you can neglect any vertical velocities acquired from gravity

For #2 you don't really need to know that it is a cylinder. By symmetry, the CM is on axis. A tapered rod that has the same mass distribution along its length would give you the same result.
 
  • #3


1. To calculate the rotational kinetic energy of the system, we need to first determine the moment of inertia of the system. In this case, we have three masses attached to the rod, so we need to use the parallel axis theorem to find the moment of inertia about the pivot point (center of mass). The moment of inertia of the rod can be calculated using the formula I = 1/12 * M * L^2. Once we have the moment of inertia, we can use the formula Krot = 1/2 * I * ω^2 to calculate the rotational kinetic energy, where ω is the angular velocity of the system. To find ω, we can use the conservation of angular momentum, since there is no external torque acting on the system. We can set the initial angular momentum (before the collision) equal to the final angular momentum (after the collision) and solve for ω. Once we have ω, we can plug it into the formula for rotational kinetic energy and solve for the answer.

2. In this problem, we need to use the formula for the center of mass, Xcm = ∫x dm / ∫dm. Since the density is non-uniform, we need to break up the cylinder into small segments and calculate the mass and center of mass for each segment. Then, we can use the integral to sum up all the masses and their respective center of mass to get the overall center of mass of the cylinder. The fact that it is a cylinder is important, as it affects the distribution of mass along the length of the cylinder. So, we cannot treat it as a uniform rod. We need to take into account the varying density along the length of the cylinder. I suggest practicing more problems with non-uniform densities to get a better understanding of how to approach them.
 
  • #4


1. For the first question, you can use the formula for rotational kinetic energy, which is 1/2 * I * ω^2. I represents the moment of inertia, which is the rotational equivalent of mass. In this case, since the rod is pivoted at its center of mass, you can use the formula for the moment of inertia of a rod pivoted at its center, which is 1/12 * M * L^2. So, the total rotational kinetic energy can be calculated as 1/2 * (1/12 * 2 * 10^2 + 1/12 * 4 * 10^2 + 1/12 * 2 * 10^2) = 50 J.

2. For the second question, you can use the formula for the center of mass of a non-uniform object, which is Xcm = ∫ x * dm / M. In this case, the mass element dm can be expressed as p(x) * dV, where p(x) is the density and dV is the volume element. Since the cylinder is uniform along the radial direction, you can use the formula for the volume of a cylinder, which is π * R^2 * dx. So, the integral can be written as Xcm = ∫ x * p(x) * π * R^2 * dx / M. You can then plug in the given values for p(x) and M to solve for Xcm.
 

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