Find the linear acceleration

1. Nov 21, 2009

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^2

For 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.

Last edited: Nov 21, 2009
2. Nov 21, 2009

tiny-tim

Hi Dark Visitor!

(have an omega: ω and try using the X2 tag just above the Reply box )
ok, now use conservation of energy (rotational KE = 1/2 Iω2).

3. Nov 21, 2009

Dark Visitor

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

4. Nov 21, 2009

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.

5. Nov 21, 2009

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?

6. Nov 22, 2009

tiny-tim

oops!

(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.

7. Nov 22, 2009

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 ac = vt2/r ?

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

8. Nov 22, 2009

tiny-tim

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 Ftotal = ma.

9. Nov 22, 2009

Dark Visitor

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

10. Nov 22, 2009

tiny-tim

Yes, of course. As I said …

11. Nov 22, 2009

Dark Visitor

How do I find T then?

12. Nov 22, 2009

tiny-tim

"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."

13. Nov 22, 2009

Dark Visitor

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

mbab = T - mr

(mb is mass of bucket, ab is accel. of the bucket, mr is mass of the reel)

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

14. Nov 22, 2009

tiny-tim

It isn't.

What are the forces on the bucket?

So what is Ftotal = ma for the bucket?

15. Nov 22, 2009

Dark Visitor

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

16. Nov 22, 2009

tiny-tim

Yup! So the equation is … ?

17. Nov 22, 2009

Dark Visitor

T = mbab

18. Nov 22, 2009

tiny-tim

What about the weight of the bucket?

19. Nov 22, 2009

Dark Visitor

That's what mb is I thought?

20. Nov 22, 2009

tiny-tim

Nooo … mb is the mass of the bucket.

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