Linear and angular acceleration of a falling cylinder? [long]

In summary: T*r.The weight points down... and we are looking at the y component of the force... that is... the component that is perpendicular to the line of the force...So the weight doesn't contribute to the torque... because it is perpendicular to the line of the force.So the net torque is:2T*r= I*alpha(remember that moment of inertia of a cylinder is 1/2mr^2)2T*r= 1/2mr^2*alphaalpha = 4T/(mr)So you have four unknowns:T, m, r, alphaBut you can relate alpha to a...in part B... in your equation... you have:
  • #1
Bob Loblaw
69
0

Homework Statement



http://img248.imageshack.us/img248/1639/cylinderjl0.jpg [Broken]

A uniform cylinder with a radius of R and mass M has been attached to two cords and the cords are wound around it and hung from the ceiling. The cylinder is released from rest and the cords unwind as the cylinder descends.
(a) draw a proper free body diagram for the cylinder;
(b) Apply Newton’s second law to the cylinder;
(c) apply Newton’s second law in rotational form to the cylinder;
(d) the two equations you have written so far
contain three unknowns; what is the relationship between the linear acceleration of the cylinder and its angular acceleration?
(e) Solve for the linear acceleration of the cylinder;
(f) What is the tension in the cords?





The attempt at a solution

A) The free body diagram should include the forces of tension pulling the cylinder up and the weight of the cylinder, right?

B) Essentially it would be Tension-weight of cylinder=angular acceleration, is that correct?

C)I know that Newton's second law in rotational form is:
Net external torque = moment of inertia x angular acceleration
So I would need to solve for moment of inertia:
I=mr^2 X Angular acceleration = net external torque.

D) So I have
Tension-weight of cylinder=angular acceleration
I=mr^2 X Angular acceleration = net external torque.

Linear Acceleration and Angular Acceleration are related by:
a. The Mass.
b. The Radius.
c. The Torque.
d. The Force.
In this problem, the mass, radius, and torque are unknown. The force working on the cylinder I believe is just gravity.

E and F) I am not sure how to do these.

I know this is a lengthy problem, but any help of guidance would be appreciately greatly! Even if ou don't know - I would settle for just a word of encouragement.
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
Bob Loblaw said:

Homework Statement



http://img248.imageshack.us/img248/1639/cylinderjl0.jpg [Broken]

A uniform cylinder with a radius of R and mass M has been attached to two cords and the cords are wound around it and hung from the ceiling. The cylinder is released from rest and the cords unwind as the cylinder descends.
(a) draw a proper free body diagram for the cylinder;
(b) Apply Newton’s second law to the cylinder;
(c) apply Newton’s second law in rotational form to the cylinder;
(d) the two equations you have written so far
contain three unknowns; what is the relationship between the linear acceleration of the cylinder and its angular acceleration?
(e) Solve for the linear acceleration of the cylinder;
(f) What is the tension in the cords?





The attempt at a solution

A) The free body diagram should include the forces of tension pulling the cylinder up and the weight of the cylinder, right?

yes.

B) Essentially it would be Tension-weight of cylinder=angular acceleration, is that correct?

not angular acceleration. The linear acceleration of the center of mass...

so write the [tex]\Sigma F_y = ma[/tex] equation...

C)I know that Newton's second law in rotational form is:
Net external torque = moment of inertia x angular acceleration
So I would need to solve for moment of inertia:
I=mr^2 X Angular acceleration = net external torque.

Yes, but the moment of inertia of the cylinder isn't mr^2... look that up... so:

net torque = I*alpha (alpha is angular acceleration)

what goes in the left side of this equation?

D) So I have
Tension-weight of cylinder=angular acceleration
I=mr^2 X Angular acceleration = net external torque.

Linear Acceleration and Angular Acceleration are related by:
a. The Mass.
b. The Radius.
c. The Torque.
d. The Force.

So in this part you need to relate a and alpha... do you know the relationship between velocity and angular velocity?

In this problem, the mass, radius, and torque are unknown. The force working on the cylinder I believe is just gravity.

E and F) I am not sure how to do these.

I know this is a lengthy problem, but any help of guidance would be appreciately greatly! Even if ou don't know - I would settle for just a word of encouragement.

Once you write your equations in parts B, C, and D... just solve them to get the answers to E and F.
 
Last edited by a moderator:
  • #3
Thanks. I made some changes and now I have:

A) My free body diagram has two tension forces up and weight pointed down.

B) Newton's 2L looks like 2T-W=ma

C) net torque = I*alpha
I=1/2mv^2 for a cylinder so:
net torque = 1/2mv^2*alpha.

I am not certain about the left side. Is net torque equal to zero?

I am not sure of the relationship between linear and angular velocity. My guess is that the radius of the cylinder will roll as it is unwound from the rope in such a way that the rotations per second will relate to the speed of the fall?
 
  • #4
Bob Loblaw said:
Thanks. I made some changes and now I have:

A) My free body diagram has two tension forces up and weight pointed down.

B) Newton's 2L looks like 2T-W=ma

yes, exactly.

C) net torque = I*alpha
I=1/2mv^2 for a cylinder so:
net torque = 1/2mv^2*alpha.

you mean:

net torque = 1/2mr^2*alpha. (not v but r).

I am not certain about the left side. Is net torque equal to zero?

What is the definition of torque? How do the tensions in the cords relate to the torque?

I am not sure of the relationship between linear and angular velocity. My guess is that the radius of the cylinder will roll as it is unwound from the rope in such a way that the rotations per second will relate to the speed of the fall?

if an object rolls without slipping... like a tire on a road... then:

v = rw

that means that

a = r*alpha
 
  • #5
By definition, torque=force x radius. The relationship between tension and torque isn't clear to me right now.
 
  • #6
Bob Loblaw said:
By definition, torque=force x radius. The relationship between tension and torque isn't clear to me right now.

yes, that's the definition. the way I think of it is... torque is

force*(perpendicular distance from the axis to the line of the force)

The force is T. Think of the axis of the cylinder... and the line of the force... draw a line segment going from the axis... to the line of the force(this line segment is just a radius of the cylinder)... the length of that line segment is r...

So torque due to one rope is T*r. Torque due to both ropes is 2T*r.
 
  • #7
Thanks, I see the problem a little better now.

So:
Newtons 2L:
2T-W=ma
Rotational form:
2T*r=1/2mr^2*alpha

The relationship between the two forms:
a = r*alpha

So we want to solve for linear acceleration. Would we take r as (1/2mr^2*alpha)/2T? If so, that would still leave me to deal with another unknown, alpha.

Thanks for your help so far. I feel fortunate that you were able to assist me.
 
  • #8
Bob Loblaw said:
Thanks, I see the problem a little better now.

So:
Newtons 2L:
2T-W=ma
Rotational form:
2T*r=1/2mr^2*alpha

The relationship between the two forms:
a = r*alpha

So we want to solve for linear acceleration. Would we take r as (1/2mr^2*alpha)/2T? If so, that would still leave me to deal with another unknown, alpha.

Thanks for your help so far. I feel fortunate that you were able to assist me.

You can cancel 1 r in:

2T*r=1/2mr^2*alpha

leaving

2T=1/2mr*alpha

so you have the 3 equations:

2T-mg=ma
2T=1/2mr*alpha
a = r*alpha

trying playing with these 3 equations... you should be able to solve for a, alpha and T in terms of m, g and r... try to go from 3 equations in 3 unknowns to 2 equations in 2 unknowns...
 
  • #9
I came up with 2r*alpha=a and mr*alpha=T

Is it even possible to have non-symbolic answers for this problem?
 
  • #10
Bob Loblaw said:
I came up with 2r*alpha=a and mr*alpha=T

Is it even possible to have non-symbolic answers for this problem?

No. you'll have to give your answer in terms of r, m and g. But your equations don't look right...

I blundered... for the first equation we should use:

mg - 2T = ma (1)... since the object is going downwards...
2T=1/2mr*alpha (2)
a = r*alpha (3)

plug in a from (3) into (1). that gives:

mg - 2T = mr*alpha ... solve this along with (2):
2T=1/2mr*alpha

solve the above 2 equations... you should be able to get T and alpha in terms of m, r and g...
 
Last edited:
  • #11
Solving for T in equation 2 gives me:
T=mr*alpha
Putting that into the equation derived from plugging (3) into (1)
mg-2mr*alpha=mr*alpha

that leaves me with mg=3 or m9.81=3 or .31kg

Am I on the right track?

The problem asks for linear acceleration of the cylinder as well as tension though I am not sure if I am any closer to finding a real solution.

EDIT: Fixed bad equation and got new results.
 
Last edited:
  • #12
Bob Loblaw said:
Solving for T in equation 2 gives me:
T=mr*alpha
Putting that into the equation derived from plugging (3) into (1)
2mr*alpha-mg=mr*alpha

that leaves me with 2-mg=1 which is mg=1. Since g is 9.81, m is 0.10kg.

Am I on the right track?

I messed up on the first equation... I modified my previous post now...

equation 2 is:

2T = 1/2mr*alpha
gives
T = 1/4 mr*alpha

plugging (3) into (1) (this is the equation I changed)

gives:

mg - 2T = mr*alpha

then plugging in the T gives:

mg - 2(1/4 mr*alpha) = mr*alpha

we can cancel m's

g - (1/2)r*alpha = r*alpha

gives (3/2)r*alpha = g

solving for alpha gives alpha = 2g/(3r)

plugging that into T = 1/4 mr*alpha

gives T = (1/4)mr*(2g)/(3r)
so T= (1/6)mg

and a = r*alpha = r*2g/(3r)
so a= 2g/3

check over this yourself... I could have blundered... look over my last post with the 3 equations again... try going through solving it and see if you get the same answers...
 
  • #13
I blundered on my algebra as well. I should have:

mg-2T=mr*alpha
T=1/4mr*alpha

mg-2(1/4)mr*alpha=mr*alpha
cancelling the mr*alpha leaves me:
mg=1/2
9.81m=1/2
m=.05kg
 
  • #14
I came out with the same answers (once I knew what to do)

so in real numbers a=2(9.81)/3=6.54
T=(1/6)mg=(1/6)(1/2)=1/12N

Is this beast finally slain?
 
  • #15
Bob Loblaw said:
I blundered on my algebra as well. I should have:

mg-2T=mr*alpha
T=1/4mr*alpha

mg-2(1/4)mr*alpha=mr*alpha
cancelling the mr*alpha leaves me:
mg=1/2
9.81m=1/2
m=.05kg

you can't cancel mr*alpha (because of the mg)... we have to divide all the terms by the same thing:

mg-2(1/4)mr*alpha=mr*alpha

if you divide each term by mr*alpha, you get

mg/(mr*alpha) -1/2 = 1
g/(r*alpha) - 1/2 = 1
 
  • #16
Bob Loblaw said:
I came out with the same answers (once I knew what to do)

so in real numbers a=2(9.81)/3=6.54
T=(1/6)mg=(1/6)(1/2)=1/12N

Is this beast finally slain?

the mg = 1/2 isn't right.
 
  • #17
You're right about my algebra being bad. So now I cannot see how mass would be solvable. If that is the case I won't get a real numerical answer. Is that normal for these types of problems?
 
  • #18
Bob Loblaw said:
You're right about my algebra being bad. So now I cannot see how mass would be solvable. If that is the case I won't get a real numerical answer. Is that normal for these types of problems?

Yeah, it is definitely normal. The idea is to solve in terms of the variables given... you're given a mass of M and radius R (we used small letters but no matter)... so your answers have to be in terms of those variables... and using constants like g is fine.

so:

alpha = 2g/(3R)
T= (1/6)Mg
so a= 2g/3

is perfectly acceptable as answers...

EDIT: If they give numbers for M and R, then of course, we'd need to get numerical answers for alpha, T and a... but without numbers, this is just fine.
 
Last edited:
  • #19
I really appreciate your help.
 
  • #20
Bob Loblaw said:
I really appreciate your help.

no prob. :)
 

1. What is linear acceleration?

Linear acceleration is the rate of change of an object's velocity in a straight line. It is typically measured in meters per second squared (m/s^2) and is a key component in understanding the motion of a falling cylinder.

2. What is angular acceleration?

Angular acceleration is the rate of change of an object's angular velocity. It is a measure of how quickly an object's rotational speed is changing and is typically measured in radians per second squared (rad/s^2).

3. How does the mass of a falling cylinder affect its linear and angular acceleration?

The mass of a falling cylinder does not affect its linear acceleration, as all objects experience the same acceleration due to gravity (9.8 m/s^2). However, the mass does affect the cylinder's moment of inertia, which in turn affects its angular acceleration. A larger mass will result in a larger moment of inertia and a slower angular acceleration.

4. How does the shape of a falling cylinder affect its linear and angular acceleration?

The shape of a falling cylinder can affect its linear and angular acceleration in several ways. A longer cylinder will experience more air resistance, which can slow down its linear acceleration. Additionally, the distribution of mass along the cylinder can affect its moment of inertia and therefore its angular acceleration. A cylinder with more mass concentrated near its ends will have a larger moment of inertia and a slower angular acceleration.

5. Can the linear and angular acceleration of a falling cylinder be calculated using equations?

Yes, the linear and angular acceleration of a falling cylinder can be calculated using various equations, such as Newton's second law of motion and the rotational equivalent of Newton's second law. These equations take into account factors such as mass, moment of inertia, and gravitational acceleration to determine the acceleration of the cylinder. However, in real-world scenarios, other factors such as air resistance and surface friction may also need to be considered.

Similar threads

Replies
13
Views
819
  • Introductory Physics Homework Help
3
Replies
97
Views
3K
  • Introductory Physics Homework Help
Replies
30
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
706
  • Introductory Physics Homework Help
Replies
33
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
238
  • Introductory Physics Homework Help
10
Replies
335
Views
7K
Replies
7
Views
236
  • Introductory Physics Homework Help
Replies
29
Views
3K
  • Introductory Physics Homework Help
Replies
10
Views
2K
Back
Top