Blocks and pulley with mass - rotational dynamics

[jsrev]

Homework Statement

Two blocks of masses m1 = 8.00 kg and m2 = 12.00 kg are connected by a string of insignificant mass. The string lies on a cylinder of mass 0.50 kg and a radius of 0.15 m. The coefficient of kinetic friction between m1 and the surface of an inclined plane is 0.10. The angle of that plane is 30°. (See figure here http://imgur.com/WtpJoye ) If the blocks were stationery at the beginning and they move 4 m, calculate:

a) The initial mechanical energy of the system.
b) The final mechanical energy of the system after traveling 4 m.
c) The velocity of the blocks after traveling 4 m.

Homework Equations

The Attempt at a Solution

For part c), I have tried the following
(W being the weight of m1)
$$EI = PEf + KEf + Wfr\\<br /> m2·g·h = m1·g·h + \frac{1}{2}·m1·v^2 + \frac{1}{2}·m2·v^2 + \frac{1}{2}·w^2 + W·sen(30°)·4·0.10$$

Which after more manipulation becomes

$$v = \sqrt{\frac{m2·g·h-m1·g·h-9.8·m1·sen(30°)·4·0.10}{\frac{1}{2}·m1 + \frac{1}{2}·m2 + \frac{1}{4}·mp}}$$
Where mp is the mass of the pulley or cylinder.
The result I get after changing the values is v = 3.73 m/s

Is this correct? Also, how can I solve the parts a) and b) ? I am not really sure about what to do next or if what I'm doing is right.

I would really appreciate any help you could give me to solve this correctly. Thanks.

 

[grzz]

Note that the KE of rotation of the cylinder is given by

$\frac{1}{2}$Iω$^{2}$ where I is the moment of inertia of the cylinder.

Also it is not clear where you have taken the reference level from which the PE$_{grav}$ is to be measured.

 

[Nathanael]

Note that the KE of rotation of the cylinder is given by

$\frac{1}{2}$Iω$^{2}$ where I is the moment of inertia of the cylinder.

He has appropriately involved the rotational inertia and rotational KE (not in his original equation, but in his final equation it appears correct.)

Also it is not clear where you have taken the reference level from which the PE$_{grav}$ is to be measured.

I don't think it matters? (Only the change in GPE is relevant)

 

[Nathanael]

$$v = \sqrt{\frac{m2·g·h-m1·g·h-9.8·m1·sen(30°)·4·0.10}{\frac{1}{2}·m1 + \frac{1}{2}·m2 + \frac{1}{4}·mp}}$$
Where mp is the mass of the pulley or cylinder.

I only see two possible problems:

First, why would it be sin(30°) and not cos(30°)? (I'm talking about this term: $9.8·m_1·sin(30°) · 4 ·0.10$)

Next, you seem to have neglected the fact that the change in height would not be the same for $m_1$ and $m_2$ (you used "h" for both of them)

By "h" you mean "4" right ?

for $m_2$ it would go down by 4 (and thus have $|ΔE_{grav}|=4m_2·g$)

but for $m_1$ it would only have $|ΔE_{grav}|=4m_1·g·sin(30°)=2·m_1·g$
(the reason is that the distance of 4 meters that it travels is the distance of the "hypotenuse" whereas the Δh traveled is only 2 meters)

 

[grzz]

My post refers to the first step of the original post.

 

[Nathanael]

My post refers to the first step of the original post.

Yes I figured, that's why I was letting you know that he did it correctly (as far as I can tell) in his final step

(so his first step was possibly a typo or something)

 

[grzz]

v = $\sqrt{(m_{2}g4 - 0.1m_{1}g4cos30 - m_{1}g4sin30)/(m_{1}/2 + m_{2}/2 + m_{p}/4)}$

as Nathanael said.

 

[jsrev]

Hi, thanks for your help. I just noticed that I didnt consider the Y-axis movement for m1 to be different from the one of m2. I didn't define PEgrav because I'm just looking for the change in energy, the problem isn't very specific on that one.

However, now I'm getting 5.3 m/s as the result of the above given equation. Can you confirm? Thank you again.

 

[Nathanael]

Hi, thanks for your help. I just noticed that I didnt consider the Y-axis movement for m1 to be different from the one of m2. I didn't define PEgrav because I'm just looking for the change in energy, the problem isn't very specific on that one.

However, now I'm getting 5.3 m/s as the result of the above given equation. Can you confirm? Thank you again.

I get 5.4 m/s

Did you take care of the sin\cos problem in the frictional term?

Or perhaps you simply forgot to round up to 5.4? (It was 5.38)

 

[jsrev]

I got 5.34 actually, I'll check out the calculations. Thanks for the help!

 

[dean barry]

post later, forgot the friction

 

[dean barry]

a)
Initial energy = 0
c + b)
A = incline angle = 30 °
g = 9.81 (m/s)/s
r = pulley radius = 0.15 metres
M = pulley mass = 0.5 kg
M1 = 8 kg
M2 = 12 kg
µ = M1 kinetic friction co-efficient = 0.1

Pulley wheel moment of inertia (i) = ½ * m * r ² = 0.005625 kg – m ²

Calculate the net force on the system :
f1 = gravitational force created by M2 = M2 * g = 117.72 N
f2 = gravitational force created by M1 = M1 * g * sine A = 39.24 N
f3 = friction force of block M1 = M1 * g * cosine A * µ = 6.7966 N

The (net) force driving the system = f1 – ( f2 + f3 ) = 71.6834 N

Equivalent mass of the pulley = 0.25 kg

Total equivalent system mass = 8 + 12 + 0.25 = 20.25 kg

System acceleration (a) :
a = net force / total equivalent mass
a = 71.6834 / 20.25
a = 3.8756 ( m / s ) / s

Block velocity (v) after 4 m :
v = square root ( u ² + ( 2 * a * s ) )
v = square root ( 2 * 3.8756 * 4 )
v = 5.3216 m/s (ANSWER c)

Final system KE :
KE (M1) = ½ * M1 * v ²
= 113.2775 Joules
KE (m2) = ½ * M2 * v ²
= 169.9163 Joules
KE pulley wheel = ½ * i * ( ( v / r ) ² )
= 3.5399 Joules

Total system final energy :
= 113.2775 + 169.9163 + 3.5399
= 286.7337 Joules (ANSWER b)

You can check answer b by subtracting (the PE gained by M2 + The work done by the friction force) from the PE translated by M2 :
= ( 12 * 9.81 * 4 ) - ( ( 8 * 9.81 * Sine A * 2 ) + ( f3 * 4 ) )
= 286.7337 Joules

Note : calcs done on excel, answers rounded on this reply.