Find the linear acceleration of points on a rolling cylinder

[vcsharp2003]

Homework Statement

[/B]
A uniform cylinder with radius r is rolling down an incline without slipping as shown in diagram below. It's angular acceleration is α and acceleration of it's center of mass C is a. What will be the linear acceleration of point of contact P and another point Q on the cylinder ? Assume that points P and Q on the cylinder lie along a circle with center as C.

I have tried to find the linear acceleration by adding linear acceleration vectors resulting from pure rotation and pure translation, but I am not sure if this approach is the right approach in this case. Or whether I am missing something.

Homework Equations

Linear acceleration resulting from pure rotation for any point on cylinder is given by cross product below.

$$\overrightarrow{a_t} = \overrightarrow{\alpha} \times \overrightarrow{r} \\ \text { where }\overrightarrow{r} \text { is the radius vector for point in question having a magnitude of } {r}$$

The Attempt at a Solution

The motion of cylinder can be considered as having pure rotation defined by angular acceleration as well as pure translation defined by acceleration of center of mass. So linear acceleration of each point will be the vector sum of acceleartions resulting from rotation and translation.

$$\therefore\overrightarrow{a_p} = \overrightarrow{\alpha} \times \overrightarrow{r_p} + \overrightarrow{a} \\
\text {and} \\ \overrightarrow{a_q} = \overrightarrow{\alpha} \times \overrightarrow{r_q} + \overrightarrow{a}$$

 

[conscience]

Hi ,

The directions of components of acceleration at points P and Q are correct . You could also find magnitude of tangential acceleration (with respect to CM)at the two points .

Since the cylinder rolls without slipping you also know the net acceleration of the point of contact P. This would give you relation between angular acceleration and acceleration of CM .

 

[kuruman]

Can you simplify the expression for $\vec{a}_P$?

 

[vcsharp2003]

Can you simplify the expression for $\vec{a}_P$?

Hi,

$$\overleftarrow{\alpha}\times\overleftarrow{r_p} =|\overleftarrow{\alpha}\rvert \; \rvert\overleftarrow{r_p}\rvert \; \sin90^\circ \text { in an upward tangential direction i.e. parallel to incline and pointing up } \\
\text {Also , } \rvert\overleftarrow{r_p}\rvert= r \text { where r is radius of cylinder} \\
\text {Since the tangent at point P is parallel to the incline, so we can simply take} \\ \text { the algebraic sum of two vectors. So, the simplified expression is as below,} \\ \text {if x-axis positive direction is down the plane } \\ \overrightarrow{a_p} =- |\overleftarrow{\alpha}\rvert \; r \; \sin90^\circ\; +|\overleftarrow{a}\rvert
$$

You can see the vector diagram for point P, where two accelerations act on point P. The first one is due to angular acceleration and is in the tangential direction at point P having a magnitude of αr and the second acceleration is parallel to the incline plane having a magnitude equal to acceleration of center of mass.

 

[vcsharp2003]

Hi ,

The directions of components of acceleration at points P and Q are correct . You could also find magnitude of tangential acceleration (with respect to CM)at the two points .

Since the cylinder rolls without slipping you also know the net acceleration of the point of contact P. This would give you relation between angular acceleration and acceleration of CM .

I think I may be missing an acceleration that points towards the center C with a magnitude of v²/r for both points P and Q. This would mean that the linear acceleration of P and Q would be a vector sum of 3 vectors. Does this sound right? This is the centripetal acceleration for a particle moving in a circle.

 

[conscience]

I think I may be missing an acceleration that points towards the center C with a magnitude of v²/r for both points P and Q. This would mean that the linear acceleration of P and Q would be a vector sum of 3 vectors. Does this sound right? This is the centripetal acceleration for a particle moving in a circle.

I believe you do not have to consider centripetal acceleration in computing the linear accelerations of P and Q .

Instead if the question had asked to calculate net acceleration of P and Q , then you need to consider centripetal component as well .

 

[kuruman]

How are $\vec{a}$ and $\ | \vec{\alpha }| r \sin90^o$ related?

You also need to be clear whether you are finding the acceleration in the reference frame of the cylinder or of the incline.

 

[haruspex]

I believe you do not have to consider centripetal acceleration in computing the linear accelerations of P and Q .

Instead if the question had asked to calculate net acceleration of P and Q , then you need to consider centripetal component as well .

That is not right.
Rotational acceleration is only meaningful for a rigid body of some extent. An individual point only has linear acceleration, and that is its net acceleration. In a steadily rotating wheel on a fixed axis, each point is accelerating towards the centre.

 

[haruspex]

I think I may be missing an acceleration that points towards the center C with a magnitude of v²/r for both points P and Q. This would mean that the linear acceleration of P and Q would be a vector sum of 3 vectors. Does this sound right? This is the centripetal acceleration for a particle moving in a circle.

Yes.
E.g. consider constant speed on the level. The linear and angular accelerations would be zero, but each point would be accelerating towards the instantaneous centre of rotation.

 

[vcsharp2003]

Yes.
E.g. consider constant speed on the level. The linear and angular accelerations would be zero, but each point would be accelerating towards the instantaneous centre of rotation.

What you said applies to point P where only the centripetal acceleration would be there as other vectors would cancel each other. But for point Q, wouldn't the linear acceleration still be the net of all three accelerations?

 

[kuruman]

Suppose I were to define the axis (or centre) of rotation as the set of points on the cylinder that are instantaneously at rest while the rest of the points on the cylinder rotate (also instantaneously) about this axis. Where would you place this axis of rotation? The cylinder rolls without slipping.

 

[vcsharp2003]

Suppose I were to define the axis (or centre) of rotation as the set of points on the cylinder that are instantaneously at rest while the rest of the points on the cylinder rotate (also instantaneously) about this axis. Where would you place this axis of rotation? The cylinder rolls without slipping.

Not sure about this answer, but I have seen in some books that all the points of contact of cylinder would be instantaneously at rest.

 

[vcsharp2003]

How are $\vec{a}$ and $\ | \vec{\alpha }| r \sin90^o$ related?

You also need to be clear whether you are finding the acceleration in the reference frame of the cylinder or of the incline.

I think these two accelerations are equal and opposite if no slipping occurs. I am using the frame of reference as the ground.

 

[kuruman]

Not sure about this answer, but I have seen in some books that all the points of contact of cylinder would be instantaneously at rest.

Consider point R on the incline directly opposite to point P on the cylinder. Rolling without slipping means that P is not moving relative to R. If the entire incline is at rest, what about the points of contact of the cylinder (instantaneously)? Do the books you have seen make sense?

 

[haruspex]

What you said applies to point P where only the centripetal acceleration would be there as other vectors would cancel each other. But for point Q, wouldn't the linear acceleration still be the net of all three accelerations?

The simplest/safest may be to revert to good old Cartesian. Taking axes parallel to and perpendicular to the slope, the centre is at (x, r) and a point on the periphery is at (x+r cos(θ), r+r sin(θ)). Differentiate twice.

 

[haruspex]

Not sure about this answer, but I have seen in some books that all the points of contact of cylinder would be instantaneously at rest.

Right, but that does not help greatly with the accelerations. The point of contact will be accelerating.

 

[vcsharp2003]

The simplest/safest may be to revert to good old Cartesian. Taking axes parallel to and perpendicular to the slope, the centre is at (x, r) and a point on the periphery is at (x+r cos(θ), r+r sin(θ)). Differentiate twice.

I do not need the exact answer in terms of x-component or y-component, but just which vectors need to be vectorially added to give the linear accelertion vector.

So, the point P will have a linear acceleration of v²/r directed towards point C since other two vectors will cancel.
For point Q, the original vectors shown in OP will be added to a linear acceleration of v²/r directed towards point C, since none of the vectors cancel any other.

 

[haruspex]

I do not need the exact answer in terms of x-component or y-component, but just which vectors need to be vectorially added to give the linear accelertion vector.

So, the point P will have a linear acceleration of v²/r directed towards point C since other two vectors will cancel.
For point Q, the original vectors shown in OP will be added to a linear acceleration of v²/r directed towards point C, since none of the vectors cancel any other.

Yes.