# Homework Help: Blocks and pulley with mass - rotational dynamics

1. Jun 10, 2014

### jsrev

1. The problem statement, all variables and given/known data

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 travelling 4 m.
c) The velocity of the blocks after travelling 4 m.

2. Relevant equations

3. The attempt at a solution

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

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.

2. Jun 10, 2014

### 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.

3. Jun 10, 2014

### Nathanael

He has appropriately involved the rotational inertia and rotational KE (not in his original equation, but in his final equation it appears correct.)
I don't think it matters? (Only the change in GPE is relevant)

4. Jun 10, 2014

### Nathanael

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 travelled is only 2 meters)

Last edited: Jun 10, 2014
5. Jun 10, 2014

### grzz

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

6. Jun 10, 2014

### Nathanael

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)

7. Jun 10, 2014

### 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.

8. Jun 10, 2014

### 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.

9. Jun 10, 2014

### Nathanael

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)

10. Jun 11, 2014

### jsrev

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

11. Jun 13, 2014

### dean barry

post later, forgot the friction

Last edited: Jun 13, 2014
12. Jun 14, 2014

### 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