# Pulley problem, acceleration, angular velocity, inertia

## 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. F=ma

## The Attempt at a Solution im on part a, and i dont know why i keep getting a=-9.81m/s/s, it should be less than that.

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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 im 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.

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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 !

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 eachother out then the gravitational forces of each mass would be working against eachother.
w1-w2=-0.7848N

If the tensions canceled eachother out then the gravitational forces of each mass would be working against eachother.
a = (m-M)g/(M+m) is right, I've edit my last post, they don't cancel each other out !,

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.

0.80082 is incorrect also.
Did you try both signs ?, try avoid rounding up as the exercice proposes,

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
Homework Helper
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.

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
Homework Helper
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.)

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
Homework Helper
.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 Vf=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?

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

If you know the final velocity you can simply say that Vf=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
Homework Helper
its average velocity.
Right. Keep going (How is average velocity related to final velocity?)

Right. Keep going (How is average velocity related to final velocity?)
final velcoity =average velocity multiplied by the time at final position so .142tf
assuming speed is linear

Nathanael
Homework Helper
final velcoity =average velocity multiplied by the time at final position so .142tf
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 ≠ Vf)

(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.)

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

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 ≠ Vf)

(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
Staff Emeritus
Homework Helper
Gold Member
v=Δx/Δt
a=(Δv/Δt)=(Δx/Δt)Δt = 0.142/(52)=.0284?
No.

Nathanael
Homework Helper
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.

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
Staff Emeritus