Rotational Energy Pulley Problem

In summary, this conversation discusses how to calculate the speed of a 4.00 kg block just before it strikes the floor, using energy methods. The pulley in the problem has a radius of 0.150 m and a moment of inertia of 0.440 kg·m2, and the rope does not slip on the pulley rim. The conversation includes equations and attempts at solving the problem, with the ultimate solution being a velocity of 2.77 m/s for the block. The problem could be solved using the Lagrangian formulation or standard Newtonian mechanics.
  • #1
SuperCass
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Homework Statement



The pulley in Fig. 9-25 has radius 0.150 m and moment of inertia 0.440 kg·m2. The rope does not slip on the pulley rim. Use energy methods to calculate the speed of the 4.00 kg block just before it strikes the floor.

Figure: http://www.webassign.net/yf9/9-25.gif

Homework Equations



Ug = mgh
Krotational = .5I

The Attempt at a Solution



I tried using torques and kinematics but that didn't work. I've tried doing mgh = .5mv^2 + .5mv^2 + .5mv^2 (each m being each block and the pulley) but that didn't work either.

Help please!

Thank you!
 
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  • #2
SuperCass said:

Homework Statement



The pulley in Fig. 9-25 has radius 0.150 m and moment of inertia 0.440 kg·m2. The rope does not slip on the pulley rim. Use energy methods to calculate the speed of the 4.00 kg block just before it strikes the floor.

Figure: http://www.webassign.net/yf9/9-25.gif

Homework Equations



Ug = mgh
Krotational = .5I

The Attempt at a Solution



I tried using torques and kinematics but that didn't work. I've tried doing mgh = .5mv^2 + .5mv^2 + .5mv^2 (each m being each block and the pulley) but that didn't work either.

Help please!

Thank you!

Ok this is a basic Atwood machine problem. There are many ways to solve it. I'll do it using the Lagrangian formulation since it's asking for energy methods, but you can do it through standard Newtonian mechanics as well (the two formalisms are physically equivalent). I'll assume that the mass on the left (the larger one) is m(1) and the mass on the right is m(2).

The kinetic energy of the system can be written as:

T = (1/2)*m(1)v^2 + (1/2)*m(2)v^2 + 1/2Iw^2

where w = v/a and w^2 = v^2/a^2.

so T = (1/2)*m(1)v^2 + (1/2)*m(2)v^2 + 1/2I*(v^2/a^2)

The potential energy of the system can be written as:

V = -m(1)gx - m(2)g(L - pi*a - x)

where L is the length of the rope, pi* a is half the circumference of the pulley, and x represents the distance of the first mass to a straight line going through the center of the pulley. So the expression L - pi*a - x represents the distance of mass m(2) to that same axis (since we want to find the potential energy associated with m(2)).

The Lagrangian is L = T - V (kinetic minus potential). Put it into the http://nerdwisdom.files.wordpress.com/2007/10/lagrange001.jpg" (where x represents q and v represents velocity, or q "dot") and you get the following equations of motion:

(m(1) + m(2) + I/a^2)*acceleration = (m(1) - m(2))*g

so the acceleration, which is constant, is:

acceleration = (m(1) - m(2))*g / (m(1) + m(2) - I/a^2)

plug in everything we know to get:

acceleration = 0.77 meters per second squared.

From here it's simple kinematics:

v(final)^2 = v(initial)^2 + 2ax

Initial velocity is 0, so that term goes away. You're left with:

v(final) = sqrt(2ax), where a is the acceleration (0.77 m/s^2) and x is the distance the 4 kg block has to travel, which is 5 meters.

Plug everything in and you should get:

v(final) = 2.77 m/s

Hope this helps.
 
Last edited by a moderator:
  • #3
Thank you so much!
 

Related to Rotational Energy Pulley Problem

What is rotational energy in a pulley system?

Rotational energy in a pulley system is the energy that is generated by the motion of the pulley. It is the product of the moment of inertia of the pulley and its angular velocity.

How do you calculate rotational energy in a pulley system?

To calculate rotational energy in a pulley system, you need to multiply the moment of inertia of the pulley by the square of its angular velocity. The equation for rotational energy is E = 1/2 * I * ω², where E is the rotational energy, I is the moment of inertia, and ω is the angular velocity.

What factors affect the rotational energy in a pulley system?

The rotational energy in a pulley system is affected by the mass and distribution of the mass of the pulley, as well as the angular velocity at which it rotates. The higher the mass and the faster the rotation, the higher the rotational energy will be.

How does rotational energy affect the efficiency of a pulley system?

Rotational energy can affect the efficiency of a pulley system by causing energy losses due to friction and heat. The more rotational energy there is, the more energy will be lost, and the less efficient the system will be.

How can rotational energy be minimized in a pulley system?

To minimize rotational energy in a pulley system, you can reduce the mass of the pulley or decrease its angular velocity. Additionally, using materials with low friction can also help reduce energy losses and increase efficiency.

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