Ratio of acceleration of two solid cylinders released on two inclined planes

[songoku]

Homework Statement

A solid cylinder is released from two inclined planes (rough surface), one has angle of 30 degree and another has angle of 60 degree (all with respect to horizontal). Find the ratio of the acceleration when the cylinder reaches the bottom of the incline plane

Relevant Equations

Newton's law
Torque
Moment of Inertia

I am not sure about my free body diagram. I assume the cylinder rolls without slipping so the forces acting on the cylinder are:

Weight directed vertically downwards
Normal force directed perpendicular to the plane
friction directed upwards, parallel to the plane

Am I correct till this point?

Thanks

 

[haruspex]

Am I correct till this point?

Yes.

 

[songoku]

Attempt 1:
∑τ = I . α (taking pivot at the center of the cylinder)
f . R = 1/2 M . R² . a / R
f = 1/2 M.a
μ.N = 1/2 M.a
μ . Mg cos θ = 1/2 .Ma
a = 2 μg cos θ → so a is proportional to cos θ

Ratio of a = cos 30^o / cos 60^o = √3Attempt 2:
∑τ = I . α (taking pivot at the point of contact of cylinder and incline plane)
M.g sin θ . R = 1/2 M . R² . a / R
g sin θ = 1/2 M.a
a = 2 g sin θ / M → so a is proportional to sin θ

Where is my mistake? Thanks

 

[haruspex]

What precisely is the equation relating a static frictional force to the corresponding normal force and the coefficient?
Check your answer before posting by considering a stationary block on a horizontal plane, no lateral forces applied.

 

[songoku]

What precisely is the equation relating a static frictional force to the corresponding normal force and the coefficient?
Check your answer before posting by considering a stationary block on a horizontal plane, no lateral forces applied.

Static friction force = 0 ≤ fs ≤ fs max

You mean I can not change the static friction into μ . N because this is for maximum static friction and the friction acting on the cylinder may not be maximum?

Thanks

 

[haruspex]

Static friction force = 0 ≤ fs ≤ fs max

You mean I can not change the static friction into μ . N because this is for maximum static friction and the friction acting on the cylinder may not be maximum?

Thanks

Exactly.
Try method 1 again, but this time don't substitute for f. Instead, write the equation for linear acceleration.

 

[songoku]

Exactly.
Try method 1 again, but this time don't substitute for f. Instead, write the equation for linear acceleration.

∑τ = I . α (taking pivot at the center of the cylinder)
f . R = 1/2 M . R2 . a / R
f = 1/2 M.a

∑F = m.a
Mg sin θ - f = M.a
Mg sin θ - 1/2 M.a = M.a
a = 2/3 g sin θ

The acceleration is proportional to sin θ but the equation is different from method 2. Where is my mistake?

Thanks

 

[haruspex]

Where is my mistake?

In your attempt 2 in post #1 you forgot about the parallel axis theorem.
You also failed to cancel an M. You should have noticed that your final equation is dimensionally inconsistent.
A good practice is to check results by considering simple special cases. In this problem, consider $\theta=\pi/2$. After cancelling the M you would have got a=2g, which is clearly not possible.

 

[songoku]

In your attempt 2 in post #1 you forgot about the parallel axis theorem.
You also failed to cancel an M. You should have noticed that your final equation is dimensionally inconsistent.
A good practice is to check results by considering simple special cases. In this problem, consider $\theta=\pi/2$. After cancelling the M you would have got a=2g, which is clearly not possible.

Ah ok I get it.

But the teacher said the friction should be directed downwards of the incline plane (not upwards)

Same as this question:

He said the friction acting on A (between the ball and table) is directed to right while I think the friction should be to the left because it opposes the motion of the ball

 

[haruspex]

the teacher said the friction should be directed downwards of the incline plane

No, definitely upwards. It reduces the linear acceleration and increases the rotational acceleration. Without friction, it would slide down with acceleration $g\sin(\theta)$ instead of $\frac 23g\sin(\theta)$.

He said the friction acting on A (between the ball and table) is directed to right while I think the friction should be to the left because it opposes the motion of the ball

In general, this case is more subtle.
Consider a round object radius R, mass m, MoI I, being pulled to the right by a force F applied at height r above its centre. Friction at tabletop is f, positive to the right.
We have $F+f=ma=mR\alpha$, $Fr-fR= I\alpha$.
Solving for f gives $f=F\frac{mrR-I}{mR^2+I}$.
[corrected from original post.]
So the direction of f depends whether $r>\frac I{mR}$. For a solid cylinder, f is to the right if $r>\frac 12R$, and no frictional force arises if r=½R.

But in your diagram r=R, so your teacher is quite correct, the frictional force is to the right. Only for a cylindrical shell is the critical height right at the top.

 

[songoku]

No, definitely upwards. It reduces the linear acceleration and increases the rotational acceleration. Without friction, it would slide down with acceleration $g\sin(\theta)$ instead of $\frac 23g\sin(\theta)$.

Is there any possible case when a solid cylinder rolls down from inclined plane the friction is downwards the incline plane?

In general, this case is more subtle.
Consider a round object radius R, mass m, MoI I, being pulled to the right by a force F applied at height r above its centre. Friction at tabletop is f, positive to the right.
We have $F+f=ma=mR\alpha$, $Fr-fR= I\alpha$.
Solving for f gives $f=F\frac{mrR-I}{mrR+I}$.

I got different result:
$F+f=ma=mR\alpha ...(1)$
$Fr-fR= I\alpha...(2)$.

Dividing (1) by (2):
$\frac{F+f}{Fr-fR}=\frac{mR}{I}$

$FI + fI =mFRr - fmR^2$

$f(I + mR^2) = mFRr - FI$

$f = F \frac{mrR-I}{mR^2+I}$ ...(3)

So the direction of f depends whether $r>\frac I{mR}$. For a solid cylinder, f is to the right if $r>\frac 12R$, and no frictional force arises if r=½R.

So if the string is connected to height 1/2 R above the center of a solid cylinder and the solid cylinder rolls without slipping, then no static friction force acting on the cylinder even though the surface is rough?

For a uniform ball, the critical height is slightly lower at $\frac 37$ above the centre.

How to get 3/7 ? I thought the moment of inertia of solid sphere is $\frac{2}{5} mR^2$ so the critical height will be 2/5 above the center of the ball? [/quote]

But in your diagram r=R, so your teacher is quite correct, the frictional force is to the right.

So the frictional force actually increases the linear acceleration and reduces rotational acceleration?

Only for a cylindrical shell is the critical height right at the top.

For cylindrical shell, the friction the friction will not be directed to the right, always to the left?I also tried the case where a force F applied at distance r below the center of an object and I got:
$f=F\frac{I-mrR}{mR^2-I}$...(4)

Questions:
1. So it will be the other way round, when the applied force F is below the center, the frictional force will be to the right when $r>\frac{I}{mR}$?

2. Plugging r = 0, I got different result for (3) and (4). Something wrong with the derivation?

3. What happens when $I = MR^2$ (for cylindrical shell case)? The denominator of (4) will be zero and what does this mean (how to interpret this)?

Thanks

 

[haruspex]

Is there any possible case when a solid cylinder rolls down from inclined plane the friction is downwards the incline plane?

Can't think of one.

I got different result:

You are right, I made a mistake in the algebra.

So if the string is connected to height 1/2 R above the center of a solid cylinder and the solid cylinder rolls without slipping, then no static friction force acting on the cylinder even though the surface is rough?

Right. If the surface were frictionless it wouid still roll without slipping.

How to get 3/7 ?

By making a silly error.

So the frictional force actually increases the linear acceleration and reduces rotational acceleration?

Yes. Without friction, the cylinder would slip so as to rotate faster but move more slowly horizontally.

For cylindrical shell, the friction the friction will not be directed to the right, always to the left?

If the applied force is at the very top then there is no friction. Anywhere below that, the frictional force is to the left.

when the applied force F is below the center

If you are now defining r as distance below the centre then you should just get equation (3) but with the sign of r switched. But why do that? Just use equation (3), allowing that r can be negative.

I got different result for (3) and (4)

(4) is wrong. You should have got $-F\frac{mRr+I}{mR^2+I}$, so f is always to the left.

 

[songoku]

If you are now defining r as distance below the centre then you should just get equation (3) but with the sign of r switched. But why do that? Just use equation (3), allowing that r can be negative.

(4) is wrong. You should have got $-F\frac{mRr+I}{mR^2+I}$, so f is always to the left.

This is what I did: let r = distance below the center, taking both F and f to the right, taking the center of the object as pivot and the object rolls without slipping

$Fr+fR=I\alpha$ ...(1)
$F+f=ma = m\alpha R$...(2)

Dividing (1) by (2):
$\frac{Fr+fR}{F+f}=\frac{I}{mR}$

$FrmR+fmR^2=FI+fI$

$f(mR^2-I)=F(I-mrR)$

$f=F\frac{I-mrR}{mR^2-I}$

But I think my case here is weird because the object speeding up to the right while rotating counter-clockwise

Edit: I think I know my mistake. Instead of $Fr+fR=I\alpha$ it should be $-Fr-fR=I\alpha$ because the rotation of the object is clockwise while those two forces produce counter-clockwise torque?

 

[haruspex]

Edit: I think I know my mistake. Instead of $Fr+fR=I\alpha$ it should be $-Fr-fR=I\alpha$ because the rotation of the object is clockwise while those two forces produce counter-clockwise torque?

Either is right, depending on which way you are taking as positive for the angular acceleration. If you take anticlockwise as positive then $Fr+fR=I\alpha$, but $a=-R\alpha$.

 

[songoku]

Either is right, depending on which way you are taking as positive for the angular acceleration. If you take anticlockwise as positive then $Fr+fR=I\alpha$, but $a=-R\alpha$.

Ah I see

I also tried for object on inclined plane.

Let the object rolls down the plane without slipping, applied force F is at r above the center and directed upwards parallel to the plane (assume it is string connected to pulley at the top of the inclined plane), there will be component of weight down the plane (W sin θ, where θ is the angle of inclined plane with respect to horizontal) and friction is downwards (parallel to the plane)

Taking the center of object as pivot:
$-Fr-fR=I\alpha$ ...(1) → the object rolls clockwise while the torque produces by F and f is counter-clockwise
$W sin \theta +f-F=m.a=m \alpha R$...(2)

Dividing (1) by (2):
$\frac{-Fr-fR}{W sin \theta +f-F}=\frac{I}{mR}$

$-FrmR-fmR^2=W sin \theta I +fI-FI$

$f(I+mR^2)=FI-W sin \theta I - FrmR$

$f=\frac{FI-W sin \theta I - FrmR}{I+mR^2}$

For f to be positive (directed downwards parallel to the inclined plane):
$FI-W sin \theta I - FrmR > 0$

$r<\frac{I(F-W sin \theta)}{FmR}$

In the case where r > 0, then F - W sin θ > 0 → F > W sin θ
But how can we know the relation between F and W sin θ? From (2) , we can not know whether W sin θ is bigger or less than F

Thanks

 

[haruspex]

But how can we know the relation between F and W sin θ?

F, W and theta are independent inputs, so you cannot in general know unless given their values.

 

[songoku]

F, W and theta are independent inputs, so you cannot in general know unless given their values.

So depending on the input, the friction can either directed upwards or downwards?

View attachment 267441

For this question, the acceleration of A is not the same as B? Acceleration of B will be twice of A?

Thanks

 

[haruspex]

So depending on the input, the friction can either directed upwards or downwards?For this question, the acceleration of A is not the same as B? Acceleration of B will be twice of A?

Thanks

Yes to both.

 

[songoku]

Yes to both.

So for question the object is on inclined plane, what is the best approach to use? Just assume the direction of the friction then calculate the tension and /or acceleration? If the tension and/or acceleration is negative, reverse the direction of friction and redo the working?

 

[haruspex]

So for question the object is on inclined plane, what is the best approach to use? Just assume the direction of the friction then calculate the tension and /or acceleration? If the tension and/or acceleration is negative, reverse the direction of friction and redo the working?

Yes, except that there is no need to redo any working. You are not really assuming the friction is in a particular direction; you are only defining a particular direction as positive for the frictional force. If you get an answer like -3N then you know it is 3N in the other direction.

 

[songoku]

Yes, except that there is no need to redo any working. You are not really assuming the friction is in a particular direction; you are only defining a particular direction as positive for the frictional force. If you get an answer like -3N then you know it is 3N in the other direction.

The direction of friction force won't affect the value of acceleration or tension? I mean maybe wrong direction of friction force will result in negative value of acceleration or tension.

View attachment 267441

For this question, if the pulley is frictionless (so no rotation on the pulley), will the acceleration of A and B be the same? or the acceleration of B will still be twice of A?

Thanks

 

[haruspex]

The direction of friction force won't affect the value of acceleration or tension? I mean maybe wrong direction of friction force will result in negative value of acceleration or tension.

The key fact is that the equation is still correct, regardless of whether the value is positive or negative. A force 1N to the right is the same as a force of -1N to the left.

For this question, if the pulley is frictionless (so no rotation on the pulley), will the acceleration of A and B be the same? or the acceleration of B will still be twice of A?

First, usually when one refers to a frictionless pulley one means that the axle is frictionless. It is generally assumed there is no slip between pulley and rope.
So I think you mean a massless pulley (so no rotational inertia) or replace the pulley with a frictionless ridge.
The relationship between the two accelerations is nothing to do with that. You could just consider a cylinder on a horizontal surface and someone walking away, pulling on a thread wrapped over the top of the cylinder. The distance the person walks will be twice the distance the cylinder rolls, so twice the speed and twice the acceleration.

 

[songoku]

Thank you very much haruspex