Solving Accelerations of A Can on A Block

[kudoushinichi88]

This problem is part of a main problem...

Homework Statement

A can is placed on a block of mass m which in turn is initially at
rest on a horizontal frictionless table. If a horizontal force F is applied to
the block, it accelerates and the cylinder rolls without slipping. Find the
linear accelerations of the block and the can with respect to the table, and the
angular acceleration of the can about its centre of mass.

Homework Equations

\tau=Fr=I\alpha

The Attempt at a Solution

linear acceleration of the block

a_{block}=\frac{F}{M+m}

linear acceleration of the can

a_{can}=0 (it is stationary with respect to the table since it is rolling without slipping)

angular acceleration of the can

from the previous part of the question, the moment of inertia of the can is

I=\frac{3}{4}MR^2

The torque is produced by the frictional force acting on the can, which equal to F. So,

FR=\frac{3}{4}MR^2\alpha

which leads to

\alpha=\frac{4F}{3MR}



Am I missing something?

 

[tiny-tim]

hi kudoushinichi88!

The torque is produced by the frictional force acting on the can, which equal to F.

nooo …

the acceleration of the can is F/(M+m), so the frictional force acting on the can is … ?

 

[kudoushinichi88]

huh, so the can is actually accelerating with respect to the table?

 

[tiny-tim]

yes, the friction is a horizontal force on the can, so it must be accelerating

(but I got the value of the acceleration wrong in my last post )

 

[kudoushinichi88]

so it's accelerating at a=F/m?

 

[tiny-tim]

are you just guessing?

work it out, using the rolling condition

 

[Delzac]

Hmmm, so it the cylinder moving with respect to the table? A bit confuse here. I suppose its position doesn't change with respect to the table, but it is accelerating in the opposing direction of the block to prevent its position, with respect to the table, from changing?

And i also have a question.

From the MIT open courseware, it the lecturer said that in rolling without slipping the velocity of centre of mass is equal to the velocity at the top of the cylinder/ circle.
But in my own textbook it says 2V_cm = V_circumference. Why is there this difference?

 

[tiny-tim]

Hmmm, so it the cylinder moving with respect to the table? A bit confuse here. I suppose its position doesn't change with respect to the table, but it is accelerating in the opposing direction of the block to prevent its position, with respect to the table, from changing?

no, you find the acceleration by calling the friction force F, and using the rolling conditon together with F = ma and Fr = Iα (so mar = Iα)

From the MIT open courseware, it the lecturer said that in rolling without slipping the velocity of centre of mass is equal to the velocity at the top of the cylinder/ circle.

(what's open courseware? is that written or a video?)

no, that's completely wrong, that's a description of sliding

 

[kudoushinichi88]

After much thinking, I realized that i should have considered the forces acting on each body separately.

Calling f the frictional force, we have acting on the block,

F-f=M_b a_b...(1)

Here M_b and a_b are masses and acceleration of the block. f here is the frictional force acted by the can ON the block!

On the can, we have

f=M_c a_c...(2)

and

fR=I\alpha...(3)

the relation for angular acceleration of the can is

a_b-a_c=R\alpha...(4)

Previously we know that the moment of inertia of the can is

I=\frac{3}{4}MR^2...(5)

The rest is just algebra.

<br /> a_b=\frac{7F}{4M_c+7M_b}

a_c=\frac{4F}{4M_c+7M_b}

and

<br /> \alpha=\frac{3F}{(4M_c+7M_b)R}

What I am not sure is if I got the angular acceleration of the can right (eqn 4). I am merely guessing. I am not sure how to explain how did I arrive at that relation. Is it correct to say that the angular acceleration here arise because of the imbalance of the accelerations in the middle of the can and at the rim of the can?

 

[Delzac]

Should the can be providing a frictional force on the block? If i were to replace the can with another block, you wouldn't say the new block is applying friction on the bottom block would you?

@tiny-tm: The velocity cm = velocity circumference came from MIT's open courseware. Meaning, free lecture. It was a series of lecture on classical mechanics that was video-ed down and posted online for everyone in the world.

This is the page.
http://ocw.mit.edu/courses/physics/8-01-physics-i-classical-mechanics-fall-1999/video-lectures/lecture-24/"

Following is the transcript:
If here is an object, the cylinder is here with radius R, and I'm going to rotate it like this and roll it in this direction, the center is called point Q.

Once it has made a complete rotation, if then the point Q has moved over a distance 2pi R, then we call that pure roll.

When we have pure roll, the velocity of this point Q, and the velocity of the circumference, if you can read that-- I'll just put a c there-- are the same.

In other words, vQ is then exactly the same as v circumference, and v circumference is always omega R.

This part always holds, but for pure roll, this holds.

 

[tiny-tim]

hi kudoushinichi88!

(sorry, I've only just noticed your reply )

yes, your proof is excellent

What I am not sure is if I got the angular acceleration of the can right (eqn 4). I am merely guessing. I am not sure how to explain how did I arrive at that relation. Is it correct to say that the angular acceleration here arise because of the imbalance of the accelerations in the middle of the can and at the rim of the can?

if in doubt, find the equation for the positions or velocities, and differentiate …

in this case, if the block moves X to the right, and the can moves c to the right relative to the block, then the can has moved x = c + X to the right, but has rotated c/R

so θ = c/R = (x - X)/R, so α = (a - A)/R

 

[kudoushinichi88]

so the block is linearly accelerating to the right with respect to the table but still rotating counter clockwise?

 

[tiny-tim]

(you mean the can )

yes, the can is accelerating slower than the block, and the only horizontal force on it is the friction, so the only torque is anti-clockwise

 

[kudoushinichi88]

woops. XD the block rotating...

Thank you very much for your help!