- #26

- 394

- 6

a=dv/dt is the acceleration not the average acceleration. dv = .142t?They are related, but they are not equal.

so than a= .142 +c

- Thread starter J-dizzal
- Start date

- #26

- 394

- 6

a=dv/dt is the acceleration not the average acceleration. dv = .142t?They are related, but they are not equal.

so than a= .142 +c

- #27

SammyS

Staff Emeritus

Science Advisor

Homework Helper

Gold Member

- 11,331

- 1,012

Most of that doesn't make sense.a=dv/dt is the acceleration not the average acceleration. dv = .142t?

so than a= .142 +c

Two previous posts told you the following:

...

the average velocity only depends on the initial and final velocities.

...

Furthermore, one of the basic kinematic equations gives you what you need very directly.The initial velocity is 0, right ? So Vavg = Vf/2

- #28

- 394

- 6

where are you getting the 2 from? shouldnt it be 5 because Vavg= change in postion/change in time.The initial velocity is 0, right ? So Vavg = Vf/2

- #29

- 224

- 35

Vavg =(Vf+Vi)/2, initial velocity is 0 so Vavg = Vf/2,

- #30

- 224

- 35

X = at^2/2, you can solve for a, Vavg = Vf+Vi/2, Vi = 0, Vavg = Vf/2

- #31

- 394

- 6

i thought Vavg=final position - initial postion / final time - initial time = (.71-0)/(5-0) so thats VVavg =(Vf+Vi)/2, initial velocity is 0 so Vavg = Vf/2,

so where is the 2 coming from?

- #32

- 224

- 35

- #33

- 394

- 6

ok i got a=.0568m/s/s

- #34

- 394

- 6

OK thanks, so let me get this straight. to solve for acceleration when given average velcocity, just use a kinematic equation; x=v0t+.5atok i got a=.0568m/s/s

- #35

- 224

- 35

- #36

- 394

- 6

yes im a noob acceleration must be constant.^{2}/2 + v_{0}t + x_{0}, to be more precise, so Δx = at^{2}/2 + v_{0}t, if you know that acceleration is constant !!

- #37

- 224

- 35

No, just keep it up !!

- #38

- 394

- 6

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=ΣmrNo, just keep it up !!

edit; i also tried the equation for a hoop I=1/2 MR

- #39

- 224

- 35

- #40

- 394

- 6

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 Iω

is radial acceleration the same as angular acceleration? it must not be

Last edited:

- #41

- 394

- 6

I dont see why this doesnt work: a

- #42

SammyS

Staff Emeritus

Science Advisor

Homework Helper

Gold Member

- 11,331

- 1,012

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.I dont see why this doesn't work: a_{r}=v^{2}r = 0.3805rad/s/s

Moreover, it's entirely irrelevant for this problem.

- #43

- 394

- 6

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?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.

- #44

Nathanael

Homework Helper

- 1,650

- 239

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 also tried the equation for a hoop I=1/2 MR^{2}

Okay, now what is the torque on 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.

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.is finding gravitational potential energy on the right track?

- #45

- 394

- 6

ok torque=FdTwo 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.

the forces acting on the pulley should be the difference of the tensions (T

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

Homework Helper

- 1,650

- 239

Yepok torque=Fd

the forces acting on the pulley should be the difference of the tensions (T_{2}-T_{1}) multiplied by distance to axis of rotation (0.053m) look right?

- #47

- 394

- 6

rotation torque is relate to torque how?Yep

edit: ok here is where im stuck, net torque = T

Last edited:

- #48

Nathanael

Homework Helper

- 1,650

- 239

You miscalculated. It should be 0.03(...) not 0.3(...)rotation torque is relate to torque how?

edit: ok here is where im stuck, T=Iα, I=T/α =(.041354/1.072)=.385760896 kg m^{2}

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

- #49

- 394

- 6

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.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

- #50

Nathanael

Homework Helper

- 1,650

- 239

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.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.

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

- Replies
- 8

- Views
- 6K

- Last Post

- Replies
- 24

- Views
- 281

- Replies
- 2

- Views
- 2K

- Last Post

- Replies
- 5

- Views
- 1K

- Last Post

- Replies
- 5

- Views
- 2K

- Last Post

- Replies
- 3

- Views
- 8K

- Last Post

- Replies
- 7

- Views
- 3K

- Last Post

- Replies
- 3

- Views
- 7K

- Last Post

- Replies
- 8

- Views
- 3K

- Replies
- 21

- Views
- 15K