Return of the Inclined Plane Double Pulley: Part II: Energy

[bjorn_v]

Homework Statement

My last inclined plane problem has a second part to it: There are some energy problems I need to solve.

A double pulley consists of two separate pieces that are welded together; a hoop of radius 10.0 cm and a disk of radius 15.0 cm. The moment of inertia of the pulley is 0.160 kg-m2. A 15.0 kg block (m1), on a 35.0° incline, is attached to the outer pulley by a massless cable and a 24.0 kg block (m2) is hanging from the inner pulley by a massless cable. The incline has a kinetic coefficient of friction of 0.250. At the instant shown in the diagram, the two blocks are at a height of 1.50 m and are moving in the appropriate direction with m2 having a speed of 0.240 m/s. The motion is considered complete when either m 1 reaches the bottom of the incline or just before m2 hits the ground.

Use energy methods to answer the following questions.
Assign the zero point reference for potential energy to be the ground for both blocks!

a) What is the speed of block m 1 when the motion is complete?
b) What is the speed of block m 2 when the motion is complete?
c) What is the total energy and the distribution of the energy of the system at the initial instant described?
d) What is the final distribution of the energy of the system just before the motion is complete?

Homework Equations

(Possibly these are relevant?)
PE = KE
mgh = 1/2 mv^2

The Attempt at a Solution

I managed to answer everything from Part I of the problem. I was able to use forces to find that m₂ moves downward, that m₁ moves up the ramp, that m₂ travels distance h (1.50m) and m₁ is pulled 2.25m. The angular acceleration of the pulley should be 8.63. Acceleration of m₁ is 1.29, m₂ is 0.863, tension in cable 1 is 133.95 and cable 2 is 214.73. Great.

I'm trying to set up some equation to find the velocities of these blocks. Unfortunately, in all the examples I'm finding there are only regular pulleys, so in this answer they solve for a single velocity.

Using the above example I've made a really crude attempt in setting up the problem (which I have little to no confidence in):

$$E_1=m_2*g*h_2-m_2*(v_2)^2/2$$
$$E_2=m_1*g*2.25*\sinθ+m_1*(v_1)^2/2$$
$$E_3=μ*m_1*g*2.25*\cosθ$$
$$E_1=E_2+E_3$$

Not even sure if using 2.25 like that would be the right way to go; I was just trying to get the x and y components of the distance m₁ travels, which I calculated in the first part, which seems relevant. But I wouldn't be able to use step 4 to set up an equation because I'd have two unknowns, v₁ and v₂, and that would be no good. And I bet the double pulley should be playing into this equation too but I'm still really confused about how that thing works.

Since the radii in my problem are different I have a very strong feeling the velocities will be different because the smaller inner pulley is going to turn faster. However I also need to factor in friction because m₁ is encountering it and is (?) going to slow down m₂.

I'd like to apply "PE = KE" which I found here, but that problem is so radically different than this one that I'm not sure how.

 

[Doc Al]

Since the radii in my problem are different I have a very strong feeling the velocities will be different because the smaller inner pulley is going to turn faster.

The key is that the inner and outer pulleys have the same angular speed and acceleration as they are welded together. Use that fact to relate the linear speeds of the two masses.

 

[Doc Al]

Also, when tracking energy changes, don't forget to include the energy of the pulley.

 

[ehild]

Homework Statement

My last inclined plane problem has a second part to it: There are some energy problems I need to solve.
a) What is the speed of block m 1 when the motion is complete?
b) What is the speed of block m 2 when the motion is complete?
c) What is the total energy and the distribution of the energy of the system at the initial instant described?
d) What is the final distribution of the energy of the system just before the motion is complete?

Homework Equations

(Possibly these are relevant?)
PE = KE
mgh = 1/2 mv^2

Wrong. The potential energy is not equal to the kinetic energy in general.
Conservation of energy states that the sum of the kinetic energy and the potential energy is constant during the motion of a system, but it is not true when there is friction.

The Attempt at a Solution

I managed to answer everything from Part I of the problem. I was able to use forces to find that m₂ moves downward, that m₁ moves up the ramp, that m₂ travels distance h (1.50m) and m₁ is pulled 2.25m. The angular acceleration of the pulley should be 8.63. Acceleration of m₁ is 1.29, m₂ is 0.863, tension in cable 1 is 133.95 and cable 2 is 214.73. Great.

I'm trying to set up some equation to find the velocities of these blocks. Unfortunately, in all the examples I'm finding there are only regular pulleys, so in this answer they solve for a single velocity.
Using the above example I've made a really crude attempt in setting up the problem (which I have little to no confidence in):

$$E_1=m_2*g*h_2-m_2*(v_2)^2/2$$
$$E_2=m_1*g*2.25*\sinθ+m_1*(v_1)^2/2$$
$$E_3=μ*m_1*g*2.25*\cosθ$$
$$E_1=E_2+E_3$$

What are those E-s?

Not even sure if using 2.25 like that would be the right way to go; I was just trying to get the x and y components of the distance m₁ travels, which I calculated in the first part, which seems relevant. But I wouldn't be able to use step 4 to set up an equation because I'd have two unknowns, v₁ and v₂, and that would be no good. And I bet the double pulley should be playing into this equation too but I'm still really confused about how that thing works.

Since the radii in my problem are different I have a very strong feeling the velocities will be different because the smaller inner pulley is going to turn faster. However I also need to factor in friction because m₁ is encountering it and is (?) going to slow down m₂.
I'd like to apply "PE = KE" which I found here, but that problem is so radically different than this one that I'm not sure how.

Why do you keep on copying the solution of a completely different problem? You feel it will not work, and it does not work. There are different accelerations, velocities and distances, all differing by the same factor R1/R2.

There is friction, you can not ignore it. In case of friction, is the mechanical energy conserved?

Anyway, you know the acceleration and the distance traveled for both blocks. It is uniformly accelerating motion. Given the path taken and the initial velocity of m2. Can you find the final velocity then?

ehild

 

[HalfLostAndSearching]

I would approach this problem by first breaking it down into smaller parts and then using energy conservation principles to solve for the desired quantities.

First, let's assign a zero point reference for potential energy to be the ground for both blocks, as stated in the problem. This means that the potential energy of both blocks at the initial instant is zero.

a) To find the speed of block m1 when the motion is complete, we can use the principle of conservation of energy. The total initial energy of the system is equal to the total final energy. At the initial instant, the only form of energy present is potential energy. At the final instant, the only form of energy present is kinetic energy. Therefore, we can equate the initial potential energy to the final kinetic energy:

m1gh = 1/2 m1v1^2

Solving for v1, we get:

v1 = √(2gh)

b) Similarly, to find the speed of block m2 when the motion is complete, we can use the same principle of conservation of energy. At the initial instant, the only form of energy present is potential energy. At the final instant, the potential energy of m2 is zero (since it has reached the ground) and the kinetic energy of m2 is 1/2 m2v2^2. Therefore, we can equate the initial potential energy to the final kinetic energy:

m2gh2 = 1/2 m2v2^2

Solving for v2, we get:

v2 = √(2gh2)

c) To find the total energy and the distribution of energy of the system at the initial instant described, we can use the equation for potential energy:

E = mgh

At the initial instant, the total energy of the system is equal to the potential energy of both blocks, which can be calculated using the above equation. The distribution of energy is as follows:

Total energy = m1gh + m2gh2
Potential energy of m1 = m1gh
Potential energy of m2 = m2gh2

d) To find the final distribution of energy of the system just before the motion is complete, we can use the same principle of conservation of energy. At the final instant, the potential energy of both blocks is zero and the kinetic energy of both blocks is 1/2m1v1^2 and 1/2