Pulley problem, acceleration, angular velocity, inertia

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  • #26
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They are related, but they are not equal.
a=dv/dt is the acceleration not the average acceleration. dv = .142t?
so than a= .142 +c
 
  • #27
SammyS
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a=dv/dt is the acceleration not the average acceleration. dv = .142t?
so than a= .142 +c
Most of that doesn't make sense.

Two previous posts told you the following:

...

the average velocity only depends on the initial and final velocities.
...
The initial velocity is 0, right ? So Vavg = Vf/2
Furthermore, one of the basic kinematic equations gives you what you need very directly.
 
  • #28
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The initial velocity is 0, right ? So Vavg = Vf/2
where are you getting the 2 from? shouldnt it be 5 because Vavg= change in postion/change in time.
 
  • #29
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Vavg =(Vf+Vi)/2, initial velocity is 0 so Vavg = Vf/2,
 
  • #30
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X = at^2/2, you can solve for a, Vavg = Vf+Vi/2, Vi = 0, Vavg = Vf/2
 
  • #31
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Vavg =(Vf+Vi)/2, initial velocity is 0 so Vavg = Vf/2,
i thought Vavg=final position - initial postion / final time - initial time = (.71-0)/(5-0) so thats Vf/5

so where is the 2 coming from?
 
  • #32
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Just work it out, you're almost there, try this because it's very simple,assuming no initial velocity if a is constant then x = 1/2 at^2 , a = 2*x/t^2, for t = 5, x = 0.071, work it out !!
 
  • #33
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Just work it out, you're almost there, try this because it's very simple,assuming no initial velocity if a is constant then x = 1/2 at^2 , a = 2*x/t^2, for t = 5, x = 0.071, work it out !!
ok i got a=.0568m/s/s
 
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  • #34
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ok i got a=.0568m/s/s
OK thanks, so let me get this straight. to solve for acceleration when given average velcocity, just use a kinematic equation; x=v0t+.5at2 and solve for t. and we use this equation because we know x, v0, and t
 
  • #35
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X = at2/2 + v0t + x0, to be more precise, so Δx = at2/2 + v0t, if you know that acceleration is constant !!
 
  • #36
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X = at2/2 + v0t + x0, to be more precise, so Δx = at2/2 + v0t, if you know that acceleration is constant !!
yes im a noob acceleration must be constant.
 
  • #37
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No, just keep it up !!
 
  • #38
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No, just keep it up !!
ok part e is getting me now. im not sure what formula to use for rotational inertial here. assuming the mass of the pulley is 0. I=Σmr2 where m is the mass of blocks and r is the radius of the pulley.

edit; i also tried the equation for a hoop I=1/2 MR2
 
  • #39
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The value of acceleration a tells us that the mass of the pully isn't 0, remember the 0.8002 you got before, that is only true when the mass is 0, so you need a another method, what about setting a relation btw α and a ?,
 
  • #40
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The value of acceleration a tells us that the mass of the pully isn't 0, remember the 0.8002 you got before, that is only true when the mass is 0, so you need a another method, what about setting a relation btw α and a ?,
i know that tangential acceleration is = rotational acceleration times radius of the pulley then just solve for α=a/r = (.0568/.053)=1.072 rad/s/s. this is incorrect, now im trying to relate the gravitational potential energy and angular velocity and using eq K=1/2 Iω2 then solve for I. kinda like my last thread.

is radial acceleration the same as angular acceleration? it must not be
 
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  • #41
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The value of acceleration a tells us that the mass of the pully isn't 0, remember the 0.8002 you got before, that is only true when the mass is 0, so you need a another method, what about setting a relation btw α and a ?,
I dont see why this doesnt work: ar=v2r = 0.3805rad/s/s
 
  • #42
SammyS
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I dont see why this doesn't work: ar=v2r = 0.3805rad/s/s
You would need to divide by r, not multiply by r, but that's just the magnitude of the radial acceleration of the cord where it's in contact with the pulley.

Moreover, it's entirely irrelevant for this problem.
 
  • #43
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You would need to divide by r, not multiply by r, but that's just the magnitude of the radial acceleration of the cord where it's in contact with the pulley.

Moreover, it's entirely irrelevant for this problem.
rotational inertial will have something to do with the Mass of the system and the radius of the pulley. is finding gravitational potential energy on the right track?
 
  • #44
Nathanael
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i also tried the equation for a hoop I=1/2 MR2
Two problems with this. (Well, zeroth, it applies to a disc not a hoop.) First, it assumes the density of the disc is uniform, which is not given. Second, you don't know the mass of the pulley.

i know that tangential acceleration is = rotational acceleration times radius of the pulley then just solve for α=a/r = (.0568/.053)=1.072 rad/s/s.
Okay, now what is the torque on the pulley?

is finding gravitational potential energy on the right track?
It's possible to solve the problem with energy considerations, (consider the change in energy after a time dt) but I would suggest you just consider the torque, instead.
 
  • #45
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Two problems with this. (Well, zeroth, it applies to a disc not a hoop.) First, it assumes the density of the disc is uniform, which is not given. Second, you don't know the mass of the pulley.


Okay, now what is the torque on the pulley?


It's possible to solve the problem with energy considerations, (consider the change in energy after a time dt) but I would suggest you just consider the torque, instead.
ok torque=Fd
the forces acting on the pulley should be the difference of the tensions (T2-T1) multiplied by distance to axis of rotation (0.053m) look right?

edit; rotational torque T=Iα I dont know T or α , but i suspect i can find α if i knew what it was beyond it being called rotational acceleration.
 
  • #46
Nathanael
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ok torque=Fd
the forces acting on the pulley should be the difference of the tensions (T2-T1) multiplied by distance to axis of rotation (0.053m) look right?
Yep
 
  • #47
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Yep
rotation torque is relate to torque how?

edit: ok here is where im stuck, net torque = T1-T2=(.23261382-.273967388)=-.041353568 but the correct answer is -0.038644208, i dont know how they got that, ive double checked my forces using 9.81m/s/s...
 
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  • #48
Nathanael
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rotation torque is relate to torque how?

edit: ok here is where im stuck, T=Iα, I=T/α =(.041354/1.072)=.385760896 kg m2
You miscalculated. It should be 0.03(...) not 0.3(...)

Anyway your answer is a bit off from rounding errors... If you were to solve it with symbols (:devil:) you would get ##I=R^2\big(\frac{g}{a}(m_2-m_1)-(m_1+m_2)\big)## which is approximately 0.036 not 0.038
 
  • #49
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You miscalculated. It should be 0.03(...) not 0.3(...)

Anyway your answer is a bit off from rounding errors... If you were to solve it with symbols (:devil:) you would get ##I=R^2\big(\frac{g}{a}(m_2-m_1)-(m_1+m_2)\big)## which is approximately 0.036 not 0.038
I did get a value of .036048701 for I but that is incorrect. Edit, nevermind i had it negative which is incorrect. 0.36048701 is correct.
 
  • #50
Nathanael
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I did get a value of .036048701 for I but that is incorrect. Edit, nevermind i had it negative which is incorrect. 0.36048701 is correct.
I solved the problem with a completely different method (the energy one) and got the same answer. I even just double checked right now by solving it with newton's laws and I get the exact same expression ##I=R^2\big(\frac{g}{a}(m_2-m_1)-(m_1+m_2)\big)##. I strongly doubt that it's incorrect.

If they want 3 significant figures then it should be 0.0361 (but I doubt they want 3 sig.figs., because the information was given only to 2 sig.figs.)

edit:
I wish I saw your edit... solved it twice for no reason :oldtongue:
 

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