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In summary, the problem at hand is a conservation of energy problem involving a cylinder and a mass attached to a rope. The system's total energy is conserved, and at two points in time, the energy can be defined as the sum of potential and kinetic energies. The first point is just before the mass is released, and the second point is just before the mass hits the ground. The kinetic energy of the cylinder must also be taken into account, which is the sum of linear and rotational kinetic energies. With this information, the problem can be solved by setting the total energy at the two points equal to each other and solving for the unknown variables.
- #1
webdivx
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Homework Statement
http://img172.imageshack.us/img172/5227/4rj8.jpg
Homework Equations
Not sure, but I think KE=.5mv2 and PE=mgh are two possibly useful equations for this problem.
The Attempt at a Solution
I have no idea how to even start this problem. I've been searching the internet for hours and I seriously need help. Step by step instructions on how to do this are really really appriciated.
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- #2
radou
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When using conservation of energy, you have to define the system which is of interest to you. You have a cylinder and a mass attached to the rope. Now, what exactly does conservation fo energy tell you, if you analyse the energies of the system at the point before the release and at the point just before the mass strikes the ground? What kind of energy do the objects in motion possess?
- #3
webdivx
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radou said:
When using conservation of energy, you have to define the system which is of interest to you. You have a cylinder and a mass attached to the rope. Now, what exactly does conservation fo energy tell you, if you analyse the energies of the system at the point before the release and at the point just before the mass strikes the ground? What kind of energy do the objects in motion possess?
they possesses Kinetic Energy if they are in motion.
- #4
radou
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webdivx said:
they possesses Kinetic Energy if they are in motion.
Exactly. Note that the cylinder possesses rotational kinetic energy, since it has an angular velocity [tex]\omega[/tex]. Find the relation between the translatoral velocity v of the other mass and between the angular velocity of the cylinder, and simply use energy conservation.
- #5
webdivx
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radou said:
Exactly. Note that the cylinder possesses rotational kinetic energy, since it has an angular velocity [tex]\omega[/tex]. Find the relation between the translatoral velocity v of the other mass and between the angular velocity of the cylinder, and simply use energy conservation.
i don't know how to do that
please walk me through
- #6
radou
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webdivx said:
i don't know how to do that
please walk me through
Okay, do you know how to use conservation of energy? The total energy of the system is constant in time, so you can pick two points in time at which you will analyse the system. Let's pick the first point as the point just before the mass on the rope is left to move freely, and the second point as the point just before the same mass falls onto the ground. Now, define the energy of the system (i.e. the sum of potential and kinetic energy) for point 1 and for point 2, and then set them equal.
- #7
webdivx
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radou said:
Okay, do you know how to use conservation of energy? The total energy of the system is constant in time, so you can pick two points in time at which you will analyse the system. Let's pick the first point as the point just before the mass on the rope is left to move freely, and the second point as the point just before the same mass falls onto the ground. Now, define the energy of the system (i.e. the sum of potential and kinetic energy) for point 1 and for point 2, and then set them equal.
KE=.5mv2 and PE=mgh
point 1 = PE
point 2 = KE + PE
so it would be mgh = .5mv2 + mgh
- #8
radou
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webdivx said:
KE=.5mv2 and PE=mgh
point 1 = PE
point 2 = KE + PEso it would be mgh = .5mv2 + mgh
There is no potential energy at point 2, since the mass is alomst on the floor. Further on, '.5mv2' (which is supposed to be 0.5mv^2, I assume), as you wrote, is the kinetic energy of the mass attached to the rope. What is the kinetic energy of the cylinder? You have to count that energy in, too.
- #9
webdivx
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radou said:
There is no potential energy at point 2, since the mass is alomst on the floor. Further on, '.5mv2' (which is supposed to be 0.5mv^2, I assume), as you wrote, is the kinetic energy of the mass attached to the rope. What is the kinetic energy of the cylinder? You have to count that energy in, too.
is the kinetic energy of the cylinder .5mv^2 = .5MR^2
- #10
Hootenanny
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The total kinetic energy of the cylinder is the sum of the linear kinetic energy and the rotational kinetic energy of the cylinder. For more information on rotational kinetic energy, http://hyperphysics.phy-astr.gsu.edu/hbase/rke.html#rke"
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- #11
webdivx
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Hootenanny said:
The total kinetic energy of the cylinder is the sum of the linear kinetic energy and the rotational kinetic energy of the cylinder. For more information on rotational kinetic energy, http://hyperphysics.phy-astr.gsu.edu/hbase/rke.html#rke"
i just don't understand. there's like a physics block in my mind. can somebody please solve for me.
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- #12
Hootenanny
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webdivx said:
i just don't understand. there's like a physics block in my mind. can somebody please solve for me.
I'm afraid we won't solve it for you here; however, if you show some effort we will guide you through the problem. Now, the total energy of the system should be conserved, the pulley is frictionless so we can write the total energy of the pulley system as;
Total energy = Gravitation Potential Energy of mass + Kinetic Energy of the mass + rotational kinetic energy of the cylinder
Can you take the next step?
Related to How Do You Solve This Energy Conservation Problem?
1. What is the conservation of energy problem?
The conservation of energy problem refers to the principle of energy conservation, which states that energy cannot be created or destroyed, but can only be transformed from one form to another. This means that the total amount of energy in a closed system remains constant over time.
2. Why is the conservation of energy important?
The conservation of energy is important because it is a fundamental law of nature and is applicable to all physical systems. It helps us understand and predict the behavior of objects and systems, and is essential for the development of many technologies.
3. How is the conservation of energy problem solved?
The conservation of energy problem is solved by using mathematical equations and principles, such as the law of conservation of energy, to analyze and calculate the energy transformations within a system. This can help us determine the initial and final states of a system and the amount of energy involved.
4. What are some real-life examples of the conservation of energy problem?
Some real-life examples of the conservation of energy problem include a swinging pendulum, a rolling ball, and a bouncing ball. In each of these cases, the total energy of the system remains constant, even as the energy is transformed from potential to kinetic and back again.
5. Are there any exceptions to the conservation of energy principle?
While the conservation of energy is a fundamental principle, there are some exceptions to its application. These include situations involving nuclear reactions, where energy can be converted into matter, and in the presence of dark energy, which is thought to cause the expansion of the universe.
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