Mechanics/Dynamics puzzle for you to chew on

[Gad-Westy]

I think I maybe am being a bit spazzy here but I can't for the life of me get my head around this. Perhaps just one of those days. Apologies in advance for the poor sketch.



The sketch shows a rotating assembly with everything rotating about a central shaft (which is vertical, i.e. the sketch shows a view looking down from above) with a speed, w1 of 400 rpm. Rigidly attached to the central shaft is an arm of fixed length, L1=0.2m. At the end of this arm is a pivot pin/shaft and mounted to this is a 'swing arm' of fixed length, L2=0.15m. The swing arm can rotate freely about the pivot point. On the other end of this swing arm, there is a roller which is free to rotate about the end of the swing arm. The roller can rotate on a shaft about the end of the swing arm. The roller and swing arm have a mass of M=10 kg. The centre of this mass is length L3=0.1m from the pivot end of the swing arm.
The roller 'swing arc' is constrained by a cylinder that is concentric about the central shaft. It has an internal radius of 0.3m. The roller will obviously run against the inside of the cylinder and so will exert a force upon it, F.
Now here's the bit I can't get my head around. First of all, how do I work out that force, F? I feel that my current approach is too simplistic. Secondly, and the real point of this question, is what is the effect on force F if L1 is varied?

[spikeyhead]

The answer is too trivial to be of interest.

[The Beaver King]

Newtons Second Law should do it

F = M x A

[WorAl]

it's the centrifugal force you need to find.

Try here

http://in.answers.yahoo.com/question/index?qid=200...

or

http://www.calctool.org/CALC/phys/newtonian/centri...

F=mV^2/r
where,
m=mass of substance
v= velocity
r = radius through which the substance is revolving

Edited by WorAl on Tuesday 18th May 15:06

[mrmr96]

Gad-Westy said:

the real point of this question, is what is the effect on force F if L1 is varied?

I think decreasing L1 will decrease F.

[spikeyhead]

There's two forces, use vectors to add them

[Gad-Westy]

WorAl said:

it's the centrifugal force you need to find.

Try here

http://in.answers.yahoo.com/question/index?qid=200...

or

http://www.calctool.org/CALC/phys/newtonian/centri...

F=mV^2/r
where,
m=mass of substance
v= velocity
r = radius through which the substance is revolving

Edited by WorAl on Tuesday 18th May 15:06

With R being the radius that the swing arms's centre of mass moves through?

[Gad-Westy]

spikeyhead said:

There's two forces, use vectors to add them

I'm not sure which forces you are referring to?

[WorAl]

Gad-Westy said:

WorAl said:

it's the centrifugal force you need to find.

Try here

http://in.answers.yahoo.com/question/index?qid=200...

or

http://www.calctool.org/CALC/phys/newtonian/centri...

F=mV^2/r
where,
m=mass of substance
v= velocity
r = radius through which the substance is revolving

Edited by WorAl on Tuesday 18th May 15:06

With R being the radius that the swing arms's centre of mass moves through?

R will be the distance to the point of contact between the rotating mass and the cylinder as that is the surface you are calculating the force on.

I will stick my neck out and say the centrifugal force exerted is 5.264kN and I think the force acting perpendicular to the cylinder wall is 3.722kN, don't quote me on that though.

Edited by WorAl on Tuesday 18th May 16:07

[blinkythefish]

Gad-Westy said:

I think I maybe am being a bit spazzy here but I can't for the life of me get my head around this. Perhaps just one of those days. Apologies in advance for the poor sketch.



The sketch shows a rotating assembly with everything rotating about a central shaft (which is vertical, i.e. the sketch shows a view looking down from above) with a speed, w1 of 400 rpm. Rigidly attached to the central shaft is an arm of fixed length, L1=0.2m. At the end of this arm is a pivot pin/shaft and mounted to this is a 'swing arm' of fixed length, L2=0.15m. The swing arm can rotate freely about the pivot point. On the other end of this swing arm, there is a roller which is free to rotate about the end of the swing arm. The roller can rotate on a shaft about the end of the swing arm. The roller and swing arm have a mass of M=10 kg. The centre of this mass is length L3=0.1m from the pivot end of the swing arm.
The roller 'swing arc' is constrained by a cylinder that is concentric about the central shaft. It has an internal radius of 0.3m. The roller will obviously run against the inside of the cylinder and so will exert a force upon it, F.
Now here's the bit I can't get my head around. First of all, how do I work out that force, F? I feel that my current approach is too simplistic. Secondly, and the real point of this question, is what is the effect on force F if L1 is varied?

It's been a while since I did this stuff. but, I suspect that you have to calculate the centripetal force for a 10KG mass rotating at 'M'. So F=m w^2 r, where r is the the distance from M to the centre shaft - you seem to have marked the roller's radius as .5m, so it should be possible to get r with a bit of geometry.

From that, you need to calculate the force at 'F', which will apply the force you calulated above, at 'M', which is a question of torque. Ff * .15 = Fm * .1, solve for Ff, given Fm in the first paragraph.

I am not 100% that this is the right method, I suspect you may have to calculate the component of F(and M) which is(are) perpendicular to the shaft....... I may need to think about this some more.

[gilla]

WorAl said:

I will stick my neck out and say the centrifugal force exerted is 5.264kN and I think the force acting perpendicular to the cylinder wall is 3.722kN, don't quote me on that though.

Edited by WorAl on Tuesday 18th May 16:07

Just quoting you on that

[Gad-Westy]

WorAl said:

Gad-Westy said:

WorAl said:

it's the centrifugal force you need to find.

Try here

http://in.answers.yahoo.com/question/index?qid=200...

or

http://www.calctool.org/CALC/phys/newtonian/centri...

F=mV^2/r
where,
m=mass of substance
v= velocity
r = radius through which the substance is revolving

Edited by WorAl on Tuesday 18th May 15:06

With R being the radius that the swing arms's centre of mass moves through?

R will be the distance to the point of contact between the rotating mass and the cylinder as that is the surface you are calculating the force on.

I will stick my neck out and say the centrifugal force exerted is 5.264kN and I think the force acting perpendicular to the cylinder wall is 3.722kN, don't quote me on that though.

Edited by WorAl on Tuesday 18th May 16:07

Cheers, Will get back to you if I get the same result with the same method.

[Gad-Westy]

blinkythefish said:

Gad-Westy said:

I think I maybe am being a bit spazzy here but I can't for the life of me get my head around this. Perhaps just one of those days. Apologies in advance for the poor sketch.



The sketch shows a rotating assembly with everything rotating about a central shaft (which is vertical, i.e. the sketch shows a view looking down from above) with a speed, w1 of 400 rpm. Rigidly attached to the central shaft is an arm of fixed length, L1=0.2m. At the end of this arm is a pivot pin/shaft and mounted to this is a 'swing arm' of fixed length, L2=0.15m. The swing arm can rotate freely about the pivot point. On the other end of this swing arm, there is a roller which is free to rotate about the end of the swing arm. The roller can rotate on a shaft about the end of the swing arm. The roller and swing arm have a mass of M=10 kg. The centre of this mass is length L3=0.1m from the pivot end of the swing arm.
The roller 'swing arc' is constrained by a cylinder that is concentric about the central shaft. It has an internal radius of 0.3m. The roller will obviously run against the inside of the cylinder and so will exert a force upon it, F.
Now here's the bit I can't get my head around. First of all, how do I work out that force, F? I feel that my current approach is too simplistic. Secondly, and the real point of this question, is what is the effect on force F if L1 is varied?

It's been a while since I did this stuff. but, I suspect that you have to calculate the centripetal force for a 10KG mass rotating at 'M'. So F=m w^2 r, where r is the the distance from M to the centre shaft - you seem to have marked the roller's radius as .5m, so it should be possible to get r with a bit of geometry.

From that, you need to calculate the force at 'F', which will apply the force you calulated above, at 'M', which is a question of torque. Ff * .15 = Fm * .1, solve for Ff, given Fm in the first paragraph.

I am not 100% that this is the right method, I suspect you may have to calculate the component of F(and M) which is(are) perpendicular to the shaft....... I may need to think about this some more.

This makes some sense. Need to work out the arc of M so will come back to you. cheers.

[WorAl]

gilla said:

WorAl said:

I will stick my neck out and say the centrifugal force exerted is 5.264kN and I think the force acting perpendicular to the cylinder wall is 3.722kN, don't quote me on that though.

Edited by WorAl on Tuesday 18th May 16:07

Just quoting you on that

Wahayyy.

Two elephants fell off a cliff....

[gilla]

what were they doing there in the 1st place?

[spikeyhead]

Gad-Westy said:

spikeyhead said:

There's two forces, use vectors to add them

I'm not sure which forces you are referring to?

There's a cenripetal force on the weight

there's also the force of the arm on the weight