Energy Methods + Pulley + Friction, finding final velocity

• RJLiberator
In summary, the problem involves a pulley with given radius and moment of inertia, a block with given mass being pulled by gravity, and another block being dragged up a ramp with given angle and coefficient of kinetic friction. The goal is to calculate the speed of the second block after the first has dropped a certain distance using energy methods. By setting the change in potential energy equal to the change in kinetic energy and factoring in the work done by friction, the final speed of the second block can be calculated. However, the change in height is not the same for both blocks and must be taken into account separately. The correct calculation results in a speed of 2.82 m/s.
RJLiberator
Gold Member

Homework Statement

The pulley in the figure has radius R=0.160m and a moment of Inertia I_p = 0.560 kg*m^2. The rope does not slip on the pulley rim. As m_4 (4kg block) drops due to gravity, m_2 (2kg block) is dragged up a ramp inclined at theta = 65 degrees. The coefficient of kinetic friction for the block/ramp is 0.100. Use energy methods to calculate the speed of m_2 after m_4 has dropped 5.00 meters. (Hint the normal force on m_2 is m_2gcos(theta) and change in height is not the same for m_2 and m_4 although they are related to each other).

The Attempt at a Solution

So first, decrease in PE is equal to increase in KE.
4*5*9.8 = 196 J.

The frictional work done is 1.96*cos65*5 = 9.8cos(65)
The force parrallel is 19.6sin(65)*5 = 98sin(65)
Final KE of the system = 196-9.8cos(65)-98sin(65)

Final KE's of the syem
M_4 = 1/2*4*v^2
M_2 = 1/2*2*v^2
Pulley = 1/2*I*w^2 ==> using w=v/R ==> 10.938v^2
Total KE = 13.938v^2

So:
13.938v^2 = 196-9.8cos(65)-98sin(65)
Which results in a 2.72 m/s

Should I have used the coefficient of friction instead of doing what I did? :O
Is there anything wrong with this reasoning?

Where's the figure?

RJLiberator
Here it is, sorry.

Attachments

• figure.png
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RJLiberator said:
Should I have used the coefficient of friction instead of doing what I did?
To calculate the work done by friction, you need the friction force. And that requires the coefficient of friction.

RJLiberator
I calculated it here The frictional work done is 1.96*cos65*5 = 9.8cos(65) which is from mass=2 gravity = 9.2 and u_k = 0.1 = 1.96.

With that in play, does the reasoning/logic of this problem hold up?
Namely: 13.938v^2 = 196-9.8cos(65)-98sin(65)

You'll have an easier time keeping things straight if you viewed it as ##\Delta E = \Delta KE + \Delta PE = Work_{friction}##. (Realizing that the work done by friction is negative, of course.)

RJLiberator
Also, when they state that the change in y is not the same for both of them, I assume this means that M_4 = 5 while m_2 you must use trigonometry as I did with 5cos(65) and 5sin(65). Correct?

RJLiberator said:
Also, when they state that the change in y is not the same for both of them, I assume this means that M_4 = 5 while m_2 you must use trigonometry as I did with 5cos(65) and 5sin(65). Correct?
Yes. The change in height is relevant when finding the change in PE. The 5sin(65) is the change in height of m_2.

RJLiberator
Hm. If I do that I insert the following:

Change in KE = final - initial
Final I found to be 13.938v_f^2
Initially the system is at rest so KEi = 0.

Change in PE
Final = 4*-5*9.8 =-196J

Force due to friction is 1.96*cos(65)*5 = 4.1416

This nets me the answer of 3.78 = v_f :O

I feel like I am doing something wrong with the change in PE ?

RJLiberator said:
Change in PE
Final = 4*-5*9.8 =-196J
That's the change in PE of the m_4 mass. What about the other mass?

RJLiberator
It would then be 2*9.81*5sin65 = 17.76
So -196+17.76 = -178.24

Plugging this back into the equation: 14*x^2-178.24=4.1416

I get x = 3.61

RJLiberator said:
It would then be 2*9.81*5sin65 = 17.76
Redo that bit of arithmetic.

RJLiberator
Oi. You are absolutely right, I had foolishly forgotten about the 5 in 5sin(65).

So 2*9.8*5sin(65) = 88.82.

So -196+88.82 = -107.18

13.938*x^2-107.18=4.1416

= 2.82 m/s which is similar to my original answer of 2.72 :O

Hm, I'm looking at both methods here and wondering why I am getting the difference of .1.

Admittedly, the way you suggested to do it was much more organized and to the point. It just made sense.
I would assume that this method would produce the more exact result, being 2.82.

RJLiberator said:
Admittedly, the way you suggested to do it was much more organized and to the point. It just made sense.
Good. It also makes it easier to spot and correct errors.

RJLiberator said:
I would assume that this method would produce the more exact result, being 2.82.
I haven't checked the arithmetic, but often such differences are due to round off errors.

RJLiberator
Excellent. Thank you for your help this morning. You clarified this question well for me.

1. How do I calculate the final velocity using energy methods and a pulley system?

The final velocity can be calculated using the equation: vf = √(2gh), where vf is the final velocity, g is the acceleration due to gravity, and h is the height difference between the starting and ending points of the object.

2. What is the role of friction in calculating final velocity in a pulley system?

Friction plays a significant role in determining the final velocity in a pulley system. It causes a loss of energy due to the conversion of kinetic energy into heat energy. Therefore, it is important to consider the effects of friction in the calculations to obtain an accurate final velocity.

3. Can energy methods be used to find final velocity in all types of pulley systems?

Yes, energy methods can be used to find the final velocity in all types of pulley systems, including single pulley, double pulley, and block and tackle systems. However, the calculation may vary depending on the complexity and specific characteristics of the pulley system.

4. How does the mass of the object affect the final velocity in a pulley system?

The mass of the object does not directly affect the final velocity in a pulley system. However, it does affect the amount of work and energy required to move the object, which can indirectly impact the final velocity. In general, objects with larger masses require more energy to move and therefore may have a lower final velocity compared to lighter objects.

5. Are there any other factors besides energy and friction that can affect the final velocity in a pulley system?

Yes, there may be other factors that can affect the final velocity in a pulley system, such as the tension in the rope, the angle of the pulley, and the efficiency of the pulley system. These factors should be taken into consideration when calculating the final velocity using energy methods and friction.

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