Pulley problem, acceleration, angular velocity, inertia

[J-dizzal]

Homework Statement

In the figure, block 1 has mass m1 = 450 g, block 2 has mass m2 = 530 g, and the pulley is on a frictionless horizontal axle and has radius R = 5.3 cm. When released from rest, block 2 falls 71 cm in 5.0 s without the cord slipping on the pulley. (a) What is the magnitude of the acceleration of the blocks? What are (b) tension T2 (the tension force on the block 2) and (c) tension T1 (the tension force on the block 1)? (d) What is the magnitude of the pulley’s angular acceleration? (e) What is its rotational inertia? Caution: Try to avoid rounding off answers along the way to the solution. Use g = 9.81 m/s2.

Homework Equations

F=ma

The Attempt at a Solution

im on part a, and i don't know why i keep getting a=-9.81m/s/s, it should be less than that.

 

[Noctisdark]

I will give a hint, what do you think of the acceleration of the two boxes, can you compare them ? can the acceleration be g ? Apply Newton's second law for both masses, T1 - m1g = m1a, T2 - m2g = m2a, Oh now I've noticed that the two acceleration have different (opposite) ditection !,

 

[J-dizzal]

I will give a hint, what do you think of the acceleration of the two boxes, can you compare them ? can the acceleration be g ? Apply Newton's second law for both masses, T1 - m1g = m1a, T2 - m2g = m2a, Oh now I've noticed that the two acceleration have different (opposite) ditection !,

a=(m₁g-m₂g)/(m₁+m₂) =-.80082m/s/s where am i going wrong?
I see that a is in opposite directions but that is the variable I am solving for. g is still 9.81 for both.

The tensions cancel each other out right?

edit; i have so far; m₁a=T-m₁g and m₂(-a)=T-m₂g I must not be doing the algebra correctly when solving for a here.

 

[Noctisdark]

The T's have the same magnitude, if the cancel them selves out, the objects would be free falling right?, the acceleration you've got seems correct to me !

 

[J-dizzal]

The T's have the same magnitude, if the cancel them selves out, the objects would be free falling right ?, you might want to check the acceleration again !,

If the tensions canceled each other out then the gravitational forces of each mass would be working against each other.
w₁-w₂=-0.7848N

 

[Noctisdark]

If the tensions canceled each other out then the gravitational forces of each mass would be working against each other.

a = (m-M)g/(M+m) is right, I've edit my last post, they don't cancel each other out !,

 

[J-dizzal]

a = (m-M)g/(M+m) is right, I've edit my last post, they don't cancel each other out !,

0.80082 is incorrect also.

 

[Noctisdark]

0.80082 is incorrect also.

Did you try both signs ?, try avoid rounding up as the exercice proposes,

 

[J-dizzal]

Did you try both signs ?, try avoid rounding up as the exercice proposes,

hehe i tried .800816327 should be positive since its asking for magnitude, but i tried negative also with no prevail.

 

[Nathanael]

When released from rest, block 2 falls 71 cm in 5.0 s without the cord slipping on the pulley.

That should be all you need to find the acceleration. (And the reasonable assumption that acceleration is constant.)

You can't find the acceleration with Newton's laws because you don't know the mass of the pulley.

 

[J-dizzal]

That should be all you need to find the acceleration. (And the reasonable assumption that acceleration is constant.)

You can't find the acceleration with Newton's laws because you don't know the mass of the pulley.

I know that its velcoity is 0.142m/s. How do you get acceleration from that? do i take the derivative of it?

 

[Nathanael]

I know that its velcoity is 0.142m/s. How do you get acceleration from that? do i take the derivative of it?

That's the velocity the whole time? Or the initial or final velocity? Or what?

Derivatives of a number are meaningless. Remember the definition of derivatives, it is a limiting process which involves the behavior around a point.
(If you only have a single point then the derivative is meaningless.)

 

[J-dizzal]

That's the velocity the whole time? Or the initial or final velocity? Or what?

Derivatives of a number are meaningless. Remember the definition of derivatives, it is a limiting process which involves the behavior around a point.
(If you only have a single point then the derivative is meaningless.)

.142 is the velocity at 5s. the velocity changes at a constant rate starting from 0m/s. if acceleration is the change in velocity over time, a=dv/dt. if i took a derivative of .142 its = 0 because its a constant. I need to express velocity as a function of time and then can take a derivative of that. its change in position is -0.71/5s
how do you find a function for velocity? only way i can think of is to graph the position vs time and get a position function then take two derivatives of that. Is that the quickest way?

 

[Nathanael]

.142 is the velocity at 5s. the velocity changes at a constant rate starting from 0m/s. if acceleration is the change in velocity over time, a=dv/dt. if i took a derivative of .142 its = 0 because its a constant. I need to express velocity as a function of time and then can take a derivative of that. its change in position is -0.71/5s

If you know the final velocity you can simply say that V_f=at (because as I said, you can assume acceleration is constant).

But 0.142 is not the final velocity. You took the total distance divided by the total time, why would that be the final velocity? What does the total distance divided by the total time represent?

 

[Noctisdark]

Aww, I've assumed that the pully is massless, If that is not provided, It would be better to follow Nathaneal's method !

 

[J-dizzal]

If you know the final velocity you can simply say that V_f=at (because as I said, you can assume acceleration is constant).

But 0.142 is not the final velocity. You took the total distance divided by the total time, why would that be the final velocity? What does the total distance divided by the total time represent?

its average velocity.

 

[Nathanael]

its average velocity.

Right. Keep going (How is average velocity related to final velocity?)

 

[J-dizzal]

Right. Keep going (How is average velocity related to final velocity?)

final velcoity =average velocity multiplied by the time at final position so .142t_f
assuming speed is linear

 

[Nathanael]

final velcoity =average velocity multiplied by the time at final position so .142t_f

No I think you're mixing something up. (The dimensions are wrong.) The final distance is the average velocity times the change in time.
(If you used symbols instead of numbers (!) then you would see at once it is not right: (Δx/Δt)Δt ≠ V_f)

(For constant acceleration,) the average velocity only depends on the initial and final velocities.

(Solving problems with symbols has another advantage, rounding errors are not exaggerated. You really should get in the habit of using symbols until finished.)

 

[Noctisdark]

The initial velocity is 0, right ? So Vavg = Vf/2

 

[J-dizzal]

No I think you're mixing something up. (The dimensions are wrong.) The final distance is the average velocity times the change in time.
(If you used symbols instead of numbers (!) then you would see at once it is not right: (Δx/Δt)Δt ≠ V_f)

(For constant acceleration,) the average velocity only depends on the initial and final velocities.

(Solving problems with symbols has another advantage, rounding errors are not exaggerated. You really should get in the habit of using symbols until finished.)

v=Δx/Δt
a=(Δv/Δt)=(Δx/Δt)Δt = 0.142/(5²)=.0284?

 

[SammyS]

v=Δx/Δt
a=(Δv/Δt)=(Δx/Δt)Δt = 0.142/(5²)=.0284?

No.

 

[Nathanael]

v=Δx/Δt
a=(Δv/Δt)=(Δx/Δt)Δt = 0.142/(5²)=.0284?

The v in the top equation is the average speed. The Δv in the bottom speed is the change in speed (a.k.a. the final speed, because it starts at rest).

You have to find a relationship between the final speed and the average speed.

 

[J-dizzal]

The v in the top equation is the average speed. The Δv in the bottom speed is the change in speed (a.k.a. the final speed, because it starts at rest).

You have to find a relationship between the final speed and the average speed.

wouldnt final speed not be related to average speed? if acceleration is constant?

 

[SammyS]

wouldnt final speed not be related to average speed? if acceleration is constant?

They are related, but they are not equal.

 

[J-dizzal]

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

 

[SammyS]

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.

 

[J-dizzal]

The initial velocity is 0, right ? So Vavg = Vf/2

where are you getting the 2 from? shouldn't it be 5 because Vavg= change in postion/change in time.

 

[Noctisdark]

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

 

[Noctisdark]

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