Falling Mass on a Pulley - Rotational Energy

In summary: Sum the torques.In summary, the conversation discusses a problem involving a frictionless pulley with a uniform solid disk shape. It provides information about the mass and radius of the pulley, as well as a 1.50 kg mass attached to a light wire. The question asks for the distance the stone must fall for the pulley to have 4.50 J of kinetic energy, as well as the percentage of the total kinetic energy that the pulley has. The conversation also mentions attempting to solve the problem using kinematics and the formula for kinetic energy, but ultimately uses an energy conservation approach to find the correct answer. The approach involves solving for the final height of the falling mass and assuming the final height of the pulley
  • #1
hitspace
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Homework Statement


A frictionless pulley has the shape of a uniform solid disk of mass 2.50 kg and radius of .2 m. A 1.50 kg mass is attached to a very light wire that is wrapped around the rim of the pulley, and the system is released from rest. a) How far must the stone fall so that the pulley has 4.50 J of kinetic energy. b) What percent of the total kinetic energy does the pulley have?

Homework Equations


KE_Rotational = (1/2)Iw^2
I_Disk = (1/2)Mr^2
v=rw
Vf^2=Vi^2 + 2gΔy

The Attempt at a Solution


Plugging in I_Disk into KE Rotational, I had a single unknown variable, w, which I solved for and got
KE = .5 [ .5(2.5)(.2)^2 ] w^2 w = 13.41 rad/s

V= rw so V=.2(13.41) = 2.683 m/s which is the linear velocity of the falling mass connected in tandem with the pulley.

My solutions manual uses some sort of conservation of energy approach at this point. I tried using
kinematics and the formula Vf^2=Vi^2 + 2gΔy as such.

2.68^2 = 0 + 2(9.8)(y) , and Delta Y = .367 m. I'm posting because my book says this answer is wrong; it says .673m. I understand that their was a energy approach to this problem, but I'm confused why my answer is not the same despite the differing approach. I didn't see any road blocks in my way, and it's making me wonder what I didn't account for trying this with kinematics.

Any help would be appreciated.
 
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  • #2
hitspace said:
2.68^2 = 0 + 2(9.8)(y)
I don't think the mass will be accelerating downwards at the same rate as if it were disconnected from the string.
 
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  • #3
Because the disk's rotational inertia will resist the acceleration you mean? Meaning acceleration would be less but unknown... which would be the roadblock to using kinematics?

So if I did this the way of energy conservation, would it go like this?

KE_rot_i + KE_i + Ug_i = KE_rot_f + KE_f + Ug_f
4.5 joule was given at the beginning of the problem.

0 + 0 + mg(h1) = (4.5 J) + .5 (1.5) (2.683)^2 + mg (h2)

h= 1.7 m ... still doesn't match. I'm guessing I screwed up again?
 
  • #4
hitspace said:
0 + 0 + mg(h1) = (4.5 J) + .5 (1.5) (2.683)^2 + mg (h2)
If you assume that h2=0, then it looks like h1 comes out to match the book's answer . . . unless my math is wrong.
 
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  • #5
No, no, you are absolutely right... though I am confused why it only works if we assume h2 = 0.
 
  • #6
Well, you can also solve for the delta, h1 - h2, which is how far it moved. It is just convenient to assign the final height to be 0.
 
  • #7
hitspace said:
0 + 0 + mg(h1) = (4.5 J) + .5 (1.5) (2.683)^2 + mg (h2)
It's often worth doing a mental approximation to check for finger trouble on the calculator.
2.7^2 is about 7; 7x.5x1.5 about 5; +4.5 =9.5; /g=1; /1.5=2/3=.67.
 
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  • #8
hitspace said:
acceleration would be less but unknown... which would be the roadblock to using kinematics?
It's not a roadblock. You can calculate the acceleration.
 
  • #9
TomHart said:
Well, you can also solve for the delta, h1 - h2, which is how far it moved. It is just convenient to assign the final height to be 0.

That is actually what I did previously - got 1.7 , and it turns out, it was calculator error. Thank you haruspex!

haruspex said:
It's not a roadblock. You can calculate the acceleration.

With regards to this, I would be interested in the general approach. Does it have to do with torque? I haven't yet reviewed that, but I'm starting that chapter now.
 
  • #10
hitspace said:
With regards to this, I would be interested in the general approach. Does it have to do with torque? I haven't yet reviewed that, but I'm starting that chapter now.
Yes.
 

FAQ: Falling Mass on a Pulley - Rotational Energy

1. How does rotational energy affect a falling mass on a pulley?

Rotational energy refers to the energy an object has due to its rotation. When a mass falls on a pulley, it causes the pulley to rotate, resulting in an increase in its rotational energy.

2. How is the rotational energy of a falling mass on a pulley calculated?

The rotational energy of a falling mass on a pulley can be calculated using the formula E = ½ Iω², where E is the rotational energy, I is the moment of inertia of the pulley, and ω is the angular velocity of the pulley.

3. What factors affect the rotational energy of a falling mass on a pulley?

The rotational energy of a falling mass on a pulley is affected by the mass of the object, the radius of the pulley, and the rotational velocity of the pulley.

4. How is the conservation of energy applied to a falling mass on a pulley?

The conservation of energy states that energy cannot be created or destroyed, only transferred from one form to another. In the case of a falling mass on a pulley, the potential energy of the falling mass is converted into rotational energy of the pulley.

5. What is the significance of understanding rotational energy in the context of a falling mass on a pulley?

Understanding rotational energy in the context of a falling mass on a pulley is important for engineers and scientists as it allows them to accurately predict the behavior and performance of pulley systems in various applications, such as elevators and cranes. It also helps in designing more efficient and effective pulley systems.

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