- #1

Anasazi

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Hello,

From what I've read in the stickies, this question may be more suited to this homework forum, although please note I'm not a student nor is this homework (it's simply for a personal project I'm doing), so please accept my apologies in advance if there are significant gaps in my knowledge that would be expected of a physics student...

I would really appreciate it if somebody could check the following for me. There's not a lot here, however I've tried to go through every step I need in detail so things are clearer hence the size:

Right, I've got a box that moves along a straight path, then once reaching the end of the path it reverses and moves back along the straight path.

The mass of the box is 5 kg.

The distance the box moves (from one side to the other) is 5 cm.

The time the box must move this distance is 0.06 seconds.

I'm trying to work out the size of the motor required to move this box within the given time constraints. Now obviously the motor is rotary and the box moves in a linear fashion, so the motor would be fixed to a crank and so would work similar to how a steam engine moves a locomotives wheel.

Just a couple of pointers, I'm ignoring any inefficiencies in the motor, gearing and friction etc. So, for the object to move, I have the following:

Distance to move object: 0.05 metres (5cm)

Mass of object: 2 KG

Time to move object: 0.06 second

Force required to move object: Mass x 9.8

Force required to move object: 2 x 9.8

Force required to move object: 19.6 Newtons

Torque required to move object: Distance (m) x Force (Newtons)

Torque required to move object: 0.05 x 19.6

Torque required to move object: 0.98 Newton-metres

RPM required to complete distance in time: 60 (seconds per minute) / time to move object

RPM required to complete distance in time: 60 / 0.06

RPM required to complete distance in time: 1000 rpm

One RPM actually moves the object twice, so halve the RPM:

RPM required to complete distance in time: 500 rpm

For the motor calculations, I have the following. Unfortunately this is all I know of the motor, so I'm unsure if I even have all the information required!

Given motor rated no-load RPM: 7000

Given motor rated input power: 650 watts

I don't know the motor efficiency, etc. but I know the motor the version down is 82% efficient, so I shall use that - I'm more interested in the process of these calculations rather than the accuracy as if need be I'll find a motor with full specs.

Calculated output power: input power x efficiency

Calculated output power: 650 x 0.82

Calculated output power: 533 watts

Horsepower: power / 745.69

Horsepower: 533 / 745.69

Horsepower: 0.71

(The braking torque formula I got from this website: <a href="http://www.elec-toolbox.com/Formulas/Motor/mtrform.htm">http://www.elec-toolbox.com/Formulas/Motor/mtrform.htm</a>)

Braking torque (lb-ft): (5252 x HP) / rpm

Braking torque (lb-ft): (5252 x 0.71) / 7000

Braking torque (lb-ft): 0.53

Braking torque (NM): lb-ft x 1.35

Braking torque (NM): 0.53 x 1.35

Braking torque (NM): 0.7155

However, this is for the motor running at the given rpm, not the required RPM. So I now need to calculate the gearing required to drop the motor rpm to the required rpm:

Gearing ratio: Motor RPM / Required RPM

Gearing ratio: 14

This means that the final braking torque is:

Braking torque (NM): Braking torque (NM) x gearing ratio

Braking torque (NM): 0.7155 x 14

Braking torque (NM): 10.02

So, as the required torque to move the object initially was 0.98NM, this motor is just over 10 times the required power.

Could somebody verify the above for me please?

Thank you.

From what I've read in the stickies, this question may be more suited to this homework forum, although please note I'm not a student nor is this homework (it's simply for a personal project I'm doing), so please accept my apologies in advance if there are significant gaps in my knowledge that would be expected of a physics student...

I would really appreciate it if somebody could check the following for me. There's not a lot here, however I've tried to go through every step I need in detail so things are clearer hence the size:

## Homework Statement

Right, I've got a box that moves along a straight path, then once reaching the end of the path it reverses and moves back along the straight path.

The mass of the box is 5 kg.

The distance the box moves (from one side to the other) is 5 cm.

The time the box must move this distance is 0.06 seconds.

I'm trying to work out the size of the motor required to move this box within the given time constraints. Now obviously the motor is rotary and the box moves in a linear fashion, so the motor would be fixed to a crank and so would work similar to how a steam engine moves a locomotives wheel.

## Homework Equations

Just a couple of pointers, I'm ignoring any inefficiencies in the motor, gearing and friction etc. So, for the object to move, I have the following:

Distance to move object: 0.05 metres (5cm)

Mass of object: 2 KG

Time to move object: 0.06 second

Force required to move object: Mass x 9.8

Force required to move object: 2 x 9.8

Force required to move object: 19.6 Newtons

Torque required to move object: Distance (m) x Force (Newtons)

Torque required to move object: 0.05 x 19.6

Torque required to move object: 0.98 Newton-metres

RPM required to complete distance in time: 60 (seconds per minute) / time to move object

RPM required to complete distance in time: 60 / 0.06

RPM required to complete distance in time: 1000 rpm

One RPM actually moves the object twice, so halve the RPM:

RPM required to complete distance in time: 500 rpm

For the motor calculations, I have the following. Unfortunately this is all I know of the motor, so I'm unsure if I even have all the information required!

Given motor rated no-load RPM: 7000

Given motor rated input power: 650 watts

I don't know the motor efficiency, etc. but I know the motor the version down is 82% efficient, so I shall use that - I'm more interested in the process of these calculations rather than the accuracy as if need be I'll find a motor with full specs.

Calculated output power: input power x efficiency

Calculated output power: 650 x 0.82

Calculated output power: 533 watts

Horsepower: power / 745.69

Horsepower: 533 / 745.69

Horsepower: 0.71

(The braking torque formula I got from this website: <a href="http://www.elec-toolbox.com/Formulas/Motor/mtrform.htm">http://www.elec-toolbox.com/Formulas/Motor/mtrform.htm</a>)

Braking torque (lb-ft): (5252 x HP) / rpm

Braking torque (lb-ft): (5252 x 0.71) / 7000

Braking torque (lb-ft): 0.53

Braking torque (NM): lb-ft x 1.35

Braking torque (NM): 0.53 x 1.35

Braking torque (NM): 0.7155

However, this is for the motor running at the given rpm, not the required RPM. So I now need to calculate the gearing required to drop the motor rpm to the required rpm:

Gearing ratio: Motor RPM / Required RPM

Gearing ratio: 14

This means that the final braking torque is:

Braking torque (NM): Braking torque (NM) x gearing ratio

Braking torque (NM): 0.7155 x 14

Braking torque (NM): 10.02

So, as the required torque to move the object initially was 0.98NM, this motor is just over 10 times the required power.

Could somebody verify the above for me please?

Thank you.

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