Calculating Speed of Block Using Energy Methods | Pulley Moment of Inertia Help

[pat666]

Homework Statement

The pulley in the diagram has a radius of 0.160 m and a moment of inertia 0.480 kg.m². The rope does not slip on the pulley rim. Use energy methods to calculate the speed of the 4.00 kg block just before it strikes the floor

Homework Equations

The Attempt at a Solution

not sure how to start this could someone give me a hand please.

 

[vela]

Start by identifying the different energies involved in the problem.

 

[pat666]

taking a stab gravitational potential and kinetic??

 

[pat666]

any more detailed help?? what is the energy in the pulley?

 

[vela]

Yes, those are the two types involved. Now write down an expression for the total kinetic energy in the system and the total potential energy of the system.

 

[pat666]

i can do most of that but i don't understand the energy for the pulley.

 

[pat666]

normally if there was no pulley it would be gph of 4kg = ke of 2kg

 

[pat666]

ok so i can figure out the mass of the pulley from i=1/2mr^2 but that wouldn't help would it \??

 

[Jolsa]

What is the kinetic energy of a rotating body?

 

[pat666]

ok can you please look at this and get back to me soonish...thanks
m_1* g*h_1+m_2 *g*h_2+1/2 I*ω^2=1/2 m_1 *v_1^2+1/2 m_2* v_2^2+1/2 I*ω^2 final
the left side is initial and the right final

 

[pat666]

i was just trying to solve that equation i wrote when i realized there are two unknown velocitys. its not ok to say that the velocity of block to will be zero at the top is it?

 

[pat666]

Just thought of something, if attached the two masses would have equal but opposite velocity's??

 

[vela]

Please put your equation between [ tex] and [ /tex] tags so it's easier to read.

Your equation isn't quite right. It's better if you approach this systematically. The total energy of the system is given by

$$E = m_1gh_1+m_2gh_2+\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2 v_2^2+\frac{1}{2}I\omega^2$$

That's just the potential energies of the two masses plus the kinetic energies of the two masses and the pulley. Now conservation of energy says that $E_i = E_f$. Each side of the equation will initially have the five terms. On the LHS, the heights and velocities will be their initial values, and on the RHS, you will have their final values. Some terms you will be able to erase because they're zero. At this point, you'll have the equation you were trying to write down earlier.

Now, as you already noted, you still have too many unknowns. Think about how the velocities of mass 1 and mass 2 are related. And how do they relate to the angular speed of the pulley?

 

[vela]

Just thought of something, if attached the two masses would have equal but opposite velocity's??

Yes! Now how is their speed v related to the angular speed ω of the pulley?

 

[pat666]

Thankyou so much! I can do it from here i think.

 

[pat666]

sorry one more thing would you take the height of m2 to be 5m at the top?? (since there different physical sizes) and just to be sure the relation of speed and angular velocity is omega =speed(v)/radius since it is not sliping??

 

[vela]

Yes and yes.

 

[pat666]

Hey did you get 3.182m/s??

 

[vela]

No, I got 2.81 m/s.

 

[vela]

What's the original equation you started from? You either started with the wrong expression or you made a bunch of algebra errors.

 

[pat666]

ok ill put it up now just wondering. if i went wron anywhere i think it was with the omega stuff i just subbed v^2/R^2 where omega^2 was

 

[pat666]

[ tex]196.2=98.1+2v-v+0.24ω^2[ /tex]

 

[pat666]

ok i am not sure how to display an equation properly sorry.

 

[pat666]

ok i figured it out myself... i can't do simple math was the problem, i hust solved my original equation with the 89 and got 2.76, that's close enough to yours to be accepted isn't it?

 

[vela]

The last term should have (v/r)^2, not v^2/r. The velocity is squared in every term, so the sign doesn't matter.

 

[pat666]

orrrrr i see now thankyou the radius is squared in my work it was just a typo... thankyou!

 

[EzaMoo]

For the life on me I can't see where I am going wrong...

196.2 = 98.1 +2V² - V² + 0.24 (V/r)²

98.1 = 2V² - V² +9.375V²

V = 3.07m/s

 

[pat666]

This for physics at cqu? your problem is that 2V^2 - V^2 +9.375V^2 is actually 2V^2 +(-V)^2 it ends up being 3V^2 + 9.375V^2

 

[EzaMoo]

Ah, I see! Yes it is. I figured from your posts you were at cqu too :)