Mass on Pulley problem need to find angular acceleration

[nickb145]

Homework Statement

A string is wrapped around a pulley of radius 3.45cm , and a weight hangs from the other end. The weight falls with a constant acceleration 3.00m/s2 .

What's the angular acceleration of the pulley?

Homework Equations

i'm writing out the equations i would think that are involved.

a=Δw/Δt
Centripital force Fr=mv^2/R ( i think)
Newtons second law
ω=ωi+at

The Attempt at a Solution

I am just so stuck...and don't know how to start.
I drew out a free body diagram and Drew out the forces for the mass and pulley


For the mass all i have is the Tension and mg. there is no weight for the pully so all i have is tension

I would need to sum the forces right?

 

[CAF123]

Homework Statement

A string is wrapped around a pulley of radius 3.45cm , and a weight hangs from the other end. The weight falls with a constant acceleration 3.00m/s2 .

What's the angular acceleration of the pulley?

Homework Equations

i'm writing out the equations i would think that are involved.

a=Δw/Δt
Centripital force Fr=mv^2/R ( i think)
Newtons second law
ω=ωi+at

The Attempt at a Solution

I am just so stuck...and don't know how to start.
I drew out a free body diagram and Drew out the forces for the mass and pulley


For the mass all i have is the Tension and mg. there is no weight for the pully so all i have is tension

I would need to sum the forces right?

Consider a portion of the string wrapped around the pulley. When the mass falls a distance x, what is the corresponding angular displacement of the pulley?

 

[nickb145]

Consider a portion of the string wrapped around the pulley. When the mass falls a distance x, what is the corresponding angular displacement of the pulley?

that would be 2∏

 

[NihalSh]

that would be 2π

No, that's not right. $2π$ means whatever distance $x$ hanging mass falls, the pulley would complete one revolution, that doesn't sound right, does it?... Think again.

$$Δθ=\frac{x}{R}$$

all need to do from here on is to differentiate it twice, then you'll have the relation between angular and linear acceleration.

 

[CAF123]

$$Δθ=\frac{x}{R}$$

all need to do from here on is to differentiate it twice, then you'll have the relation between angular and linear acceleration.

Yes, correct.

 

[nickb145]

No, that's not right. $2π$ means whatever distance $x$ hanging mass falls, the pulley would complete one revolution, that doesn't sound right, does it?... Think again.

$$Δθ=\frac{x}{R}$$

all need to do from here on is to differentiate it twice, then you'll have the relation between angular and linear acceleration.

Akk.. I didn't read that correctly

 

[nickb145]

SO if i understand this To get the angular displacement i need to take the constant acceleration and divide it by radius (in meteres .0345) then i can find angular acceleration?

Sorry for the stupid questions :(

 

[NihalSh]

SO if i understand this To get the angular displacement i need to take the constant acceleration and divide it by radius (in meteres .0345) then i can find angular acceleration?

Sorry for the stupid questions :(

Yes!

$$a=αR$$

 

[nickb145]

Yes!

$$a=αR$$

Ok!

So according to my book Angular acceleration is

α= Δω/ΔT I would think that wouldn't be right because i don't know the time.

but aside from my own thoughts and use a=αR is R the Angular displacement? If so, multiply it by the acceleration. then convert it to rad/s^2

If not, please correct me.

 

[nickb145]

that makes no sense, hold on let me think it out. i need to find the angular velocity i think...but how do i find the time...

 

[NihalSh]

Ok!

So according to my book Angular acceleration is

α= Δω/ΔT I would think that wouldn't be right because i don't know the time.

but aside from my own thoughts and use a=αR is R the Angular displacement? If so, multiply it by the acceleration. then convert it to rad/s^2

If not, please correct me.

Your book is perfectly right. R is, in this case, radius of pulley.

$$ω=\frac{v}{R}$$
$$α= \frac{Δω}{ΔT}=\frac{1}{R}*\frac{Δv}{ΔT}=lim_{ΔT→0} (\frac{1}{R}*\frac{Δv}{ΔT})=\frac{1}{R}*(lim_{ΔT→0} (\frac{Δv}{ΔT}))=\frac{a}{R}$$

that makes no sense, hold on let me think it out. i need to find the angular velocity i think...but how do i find the time...

Your question asked you to find angular acceleration, there is no mention of angular velocity!

 

[nickb145]

Your book is perfectly right. R is, in this case, radius of pulley.

$$ω=\frac{v}{R}$$
$$α= \frac{Δω}{ΔT}=\frac{1}{R}*\frac{Δv}{ΔT}=lim_{ΔT→0} (\frac{1}{R}*\frac{Δv}{ΔT})=\frac{1}{R}*(lim_{ΔT→0} (\frac{Δv}{ΔT}))=\frac{a}{R}$$



Your question asked you to find angular acceleration, there is no mention of angular velocity!

you're right. just thinking 'out loud' i guess.

sorry I'm kind of confused by this: could you explain it? My mind will probably kick into gear
$$α= \frac{Δω}{ΔT}=\frac{1}{R}*\frac{Δv}{ΔT}=lim_{ΔT→0} (\frac{1}{R}*\frac{Δv}{ΔT})=\frac{1}{R}*(lim_{ΔT→0} (\frac{Δv}{ΔT}))=\frac{a}{R}$$

 

[NihalSh]

you're right. just thinking 'out loud' i guess.

sorry I'm kind of confused by this: could you explain it? My mind will probably kick into gear
$$α= \frac{Δω}{ΔT}=\frac{1}{R}*\frac{Δv}{ΔT}=lim_{ΔT→0} (\frac{1}{R}*\frac{Δv}{ΔT})=\frac{1}{R}*(lim_{ΔT→0} (\frac{Δv}{ΔT}))=\frac{a}{R}$$

look in the original post I mentioned, $θ=\frac{x}{R}$...this is the definition of angle in radians. $x$ is the length of arc of the given circle. Like I said previously, differentiating it twice would give you the required relation $a=αR$. You could view the above calculations as following:
$$Δθ=\frac{Δx}{R}$$
$$ω_{average}=\frac{Δθ}{Δt}=\frac{1}{R}*\frac{Δx}{Δt}$$
$$ω_{instantaneous}=lim_{Δt→0} (\frac{Δθ}{Δt})=lim_{Δt→0} (\frac{1}{R}*\frac{Δx}{Δt})=\frac{1}{R}*(lim_{Δt→0} (\frac{Δx}{Δt}))=\frac{dθ}{dt}=\frac{1}{R}*\frac{dx}{dt}=\frac{v}{R}$$
$$Δω=\frac{Δv}{R}$$
$$α_{average}= \frac{Δω}{Δt}=\frac{1}{R}*\frac{Δv}{Δt}$$
$$α_{instantaneous}= lim_{Δt→0} (\frac{Δω}{Δt})=lim_{Δt→0} (\frac{1}{R}*\frac{Δv}{Δt})=\frac{1}{R}*(lim_{Δt→0} (\frac{Δv}{Δt}))=\frac{dω}{dt}=\frac{1}{R}*\frac{dv}{dt}=\frac{a}{R}$$

$lim$ are used to define derivatives. That's all there is to it. I have showed it in both notations, first with limits and then slightly on the right with $\frac{d}{dt}$ notation. I am assuming you have studied basic calculus, otherwise I wouldn't be talking about limits and derivatives.

If you could be a little bit more clear maybe I could clear your doubt, what is it that confuses you about these equations?

 

[nickb145]

Ak sorry! But yes i was thinking of the other question that my HW was asking me. Which in this case was asking for the angular velocity of the pully when the Mass was 1.3m above the ground.

But i have the angular acceleration, and thank you for the assistance!
But I'm still trying to figure out the angular velocity

 

[nickb145]

Ak sorry! But yes i was thinking of the other question that my HW was asking me. Which in this case was asking for the angular velocity of the pully when the Mass was 1.3m above the ground.

But i have the angular acceleration, and thank you for the assistance!
But I'm still trying to figure out the angular velocity

EDIT: I think i found the time 1.3=gt²/2 which is t=.5150787
now i plug it into ω=(angular acceleration) x Time and i get 44.81 but it says that it isn't right.

suggestions?

 

[NihalSh]

EDIT: I think i found the time 1.3=gt²/2 which is t=.5150787
now i plug it into ω=(angular acceleration) x Time and i get 44.81 but it says that it isn't right.

suggestions?

how did you find time? and time for what?...I think you haven't mentioned some remaining part of the question. Because according to the information in the original post, there is no way and no need to find time. and even if there is finding time for what?

 

[nickb145]

I needed to find angular velocity. But i did figure it out

 

[NihalSh]

I needed to find angular velocity. But i did figure it out

It's always better to mention the whole question for completeness. I can't help any further in solving for angular velocity, without knowing the complete question. But since you have figured it out, its no problem.

 

[nickb145]

No problem, thank you for the help though.

I sort of mentioned the second question as "Which in this case was asking for the angular velocity of the pully when the Mass was 1.3m above the ground."

 

[NihalSh]

No problem, thank you for the help though.

I sort of mentioned the second question as "Which in this case was asking for the angular velocity of the pully when the Mass was 1.3m above the ground."

There should be initial height given to solve for that.

 

[nickb145]

the initial height is 1.3 final is 0. It asks to find the angular velocity of the pully when the block hits the ground. Thats at least how i solved it

 

[NihalSh]

the initial height is 1.3 final is 0. It asks to find the angular velocity of the pully when the block hits the ground. Thats at least how i solved it

you can find the velocity of block since acceleration is given. Then relating angular and linear speed, you can solve for angular speed.

Assuming initial velocity is zero:
$$v=\sqrt{2.a.Δx}$$
$$v=ω.R$$

Edit: There is absolutely no need to solve for time!