- #36
J-dizzal
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- 6
yes I am a noob acceleration must be constant.Noctisdark said:X = at2/2 + v0t + x0, to be more precise, so Δx = at2/2 + v0t, if you know that acceleration is constant !
yes I am a noob acceleration must be constant.Noctisdark said:X = at2/2 + v0t + x0, to be more precise, so Δx = at2/2 + v0t, if you know that acceleration is constant !
ok part e is getting me now. I am 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.Noctisdark said:No, just keep it up !
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 I am 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.Noctisdark said: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 don't see why this doesn't work: ar=v2r = 0.3805rad/s/sNoctisdark said: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 ?,
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.J-dizzal said:I don't see why this doesn't work: ar=v2r = 0.3805rad/s/s
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?SammyS said: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.
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.J-dizzal said:i also tried the equation for a hoop I=1/2 MR2
Okay, now what is the torque on the pulley?J-dizzal said: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.
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.J-dizzal said:is finding gravitational potential energy on the right track?
ok torque=FdNathanael said: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.
YepJ-dizzal said: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?
rotation torque is relate to torque how?Nathanael said:Yep
You miscalculated. It should be 0.03(...) not 0.3(...)J-dizzal said:rotation torque is relate to torque how?
edit: ok here is where I am stuck, T=Iα, I=T/α =(.041354/1.072)=.385760896 kg m2
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.Nathanael said: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 () 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 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.J-dizzal said: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.