What is the linear acceleration of the bucket?

[Dark Visitor]

A 5 kg cylindrical reel with a radius of .6 m and a frictionless axle starts from rest and speeds up uniformly as a 3 kg bucket falls into a well, making a light rope unwind from the reel. The moment of inertia of a cylinder is I = 1/2MR^2. The linear acceleration of the bucket is:

a. 9.8 m/s^2
b. 6.3 m/s^2
c. 5.4 m/s^2
d. 3.7 m/s^2For the inertia for the cylinder, I got:
I = 1/2(5 kg)(.6 m)^2 = .9 kg*m^2

I don't know what to do next though. Any help would be appreciated. Please go into detail with every step so I understand what you mean, as I need to show my steps and work. Thanks.

 

[tiny-tim]

Hi Dark Visitor!

(have an omega: ω and try using the X² tag just above the Reply box )

For the inertia for the cylinder, I got:
I = 1/2(5 kg)(.6 m)^2 = .9 kg*m^2 …

ok, now use conservation of energy (rotational KE = 1/2 Iω²).

 

[Dark Visitor]

But we don't know what ω is. Does that mean I have to go find that?

 

[tiny-tim]

Use v = dx/dt and ω = v/radius (btw, is the radius 6 or 0.6?), and you'll get an equation with dx/dt and x.

 

[Dark Visitor]

My mistake. It is .600 m. I will be sure to correct that. Thanks for pointing that out.

What does the d stand for in the equation v = dx/dt?

 

[tiny-tim]

oops!

My mistake. It is .600 m. I will be sure to correct that. Thanks for pointing that out.

What does the d stand for in the equation v = dx/dt?

(just got up :zzz: …)

oops!

silly me … you haven't done calculus yet!

ok, complete change of plan …

call the linear acceleration a (so the angular acceleration is a/r), and the tension T,

then do good ol' Newton's second law on the bucket, and his rotational version on the reel …

that gives you two equations in two unknowns (T and a), so eliminate T to get a.

 

[Dark Visitor]

You got up at 3 a.m.? Either that, or you live in another time zone lol.

Anyways, I'm confused. Isn't linear acceleration's equation a_c = v_t²/r ?

And I'm unsure of how to do the Newton's Second Law. Is there an equation?

 

[tiny-tim]

Isn't linear acceleration's equation a_c = v_t²/r ?

And I'm unsure of how to do the Newton's Second Law. Is there an equation?

No, that's centripetal acceleration … linear acceleration is ordinary acceleration (the question asks for the linear acceleration of the bucket).

Newton's second law is F_total = ma.

 

[Dark Visitor]

Does that mean I have to find T (the tension)? And if so, how would I do that?

 

[tiny-tim]

Does that mean I have to find T (the tension)? And if so, how would I do that?

Yes, of course. As I said …

call the linear acceleration a (so the angular acceleration is a/r), and the tension T,

then do good ol' Newton's second law on the bucket, and his rotational version on the reel …

that gives you two equations in two unknowns (T and a), so eliminate T to get a.

 

[Dark Visitor]

How do I find T then?

 

[tiny-tim]

How do I find T then?

"do good ol' Newton's second law on the bucket, and his rotational version on the reel …

that gives you two equations in two unknowns (T and a), so eliminate T to get a."

 

[Dark Visitor]

When I set it up in equation form, I get:

m_ba_b = T - m_r

(m_b is mass of bucket, a_b is accel. of the bucket, m_r is mass of the reel)

But I feel like I am doing everything wrong. I don't think that equation is right.

 

[tiny-tim]

It isn't.

What are the forces on the bucket?

So what is F_total = ma for the bucket?

 

[Dark Visitor]

On the bucket, there is only tension in the rope and weight of the bucket.

 

[tiny-tim]

Yup! So the equation is … ?

 

[Dark Visitor]

T = m_ba_b

 

[tiny-tim]

On the bucket, there is only tension in the rope and weight of the bucket.

T = m_ba_b

What about the weight of the bucket?

 

[Dark Visitor]

That's what m_b is I thought?

 

[tiny-tim]

That's what m_b is I thought?

Nooo … m_b is the mass of the bucket.

You need the weight (as part of the F in F = m_ba_b)

 

[Dark Visitor]

so the equation is:

T + m_bg = m_ba_b

Is that correct?

 

[tiny-tim]

so the equation is:

T + m_bg = m_ba_b

Is that correct?

Not quite.

 

[Dark Visitor]

Where am I going wrong? I am getting more and more confused...

 

[tiny-tim]

T is up, mg is down.

 

[Dark Visitor]

so instead of what I had before, the equation is:

T - m_bg = m_ba_b ?

But after that, what do I do? I don't know tension or the acceleration.

 

[tiny-tim]

so instead of what I had before, the equation is:

T - m_bg = m_ba_b ?

That's right!

But after that, what do I do? I don't know tension or the acceleration.

Now you do the same thing for the reel, using τ = Iα instead of F = ma.

 

[Dark Visitor]

but in your last equation, is T the tension, or is it torque? and how do I know what $$\alpha$$ is?

 

[tiny-tim]

(copy-and-paste my α)

Yes, T is tension, and τ is torque. And α = a_b/r.

 

[Dark Visitor]

Okay, that makes sense so far. SO now I have:

τ = (.9 kg*m²)(a_b/ .6 m)

But we still can't find torque or the acceleration because there are 2 unknowns in the new equation...

 

[tiny-tim]

But we still can't find torque or the acceleration because there are 2 unknowns in the new equation...

i] what is τ in terms of T?

ii] you have the original equation also