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Anonymous Poster #1

torque required for tilting

08/30/2013 1:12 AM

weight to be tilted = 4200kg

time = 2mins

angle of rotation 360 degree

motor speed = 1000rpm
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lyn
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#1

Re: torque required for tilting

08/30/2013 1:20 AM

DO YOUR OWN HOMEWORK!
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dineshjegdeesh
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#2
In reply to #1

Re: torque required for tilting

08/30/2013 1:30 AM

got it.... just want to confirm........

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nick name
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#6
In reply to #2

Re: torque required for tilting

08/30/2013 4:20 AM

Show your results and we will tell if they are good or bad.
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Anonymous Poster #1
#7
In reply to #6

Re: torque required for tilting

08/30/2013 4:26 AM

got stuck,,,, coz it is an L shaped object to be tilted
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PWSlack
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#8
In reply to #7

Re: torque required for tilting

08/30/2013 4:34 AM

This object? If so, then a duplicate thread report is due.

If the tilt axis passes through the centre-of-gravity of the assembly, then no torque is needed other than that to overcome friction at the bearing and wind resistance.

With such little information to go on, based on a seeming reluctance to show the calculations carried out so far, the other imminent possibility for this thread is a homework report.

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LongintheTooth
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#3

Re: torque required for tilting

08/30/2013 1:32 AM

If you run the axis of tilt though the center of gravity the torque requirement will be minimal.;)

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Tornado
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#4

Re: torque required for tilting

08/30/2013 3:39 AM

Insufficient information.

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Anonymous Poster #1
#5
In reply to #4

Re: torque required for tilting

08/30/2013 4:07 AM

information required??????
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PWSlack
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#9
In reply to #5

Re: torque required for tilting

08/30/2013 4:36 AM

Well, how about the calulation details so far and an indication where the stiction in them is coming from?

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dineshjegdeesh
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#10
In reply to #9

Re: torque required for tilting

08/30/2013 4:42 AM

just calculated the rpm required

how to calculate cg for L shaped objects

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PWSlack
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#11
In reply to #10

Re: torque required for tilting

08/30/2013 5:05 AM

No calculations for review, then?

At the CoG, the sum of all the turning moments is zero.

This post is millimetres from having a homework report raised.

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Tornado
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#12
In reply to #5

Re: torque required for tilting

08/30/2013 5:13 AM

The center of rotation and the location of the object's CG in relation to this center.

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JBTardis
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#13

Re: torque required for tilting

08/30/2013 7:54 AM

FYI the unit 'kg' should be used with mass, not weight. For weight one should use a force unit (pounds, Newtons, etc.).

Sorry....just one of my pet peeves.

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LongintheTooth
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#14
In reply to #13

Re: torque required for tilting

08/30/2013 11:36 AM

No accounting for the education or lack thereof in others. Some very bright people fail to use proper terms, and the english language does not help. Then again, the speed of light is greater than the speed of sound. That's why some people appear bright until you hear them speak.

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LongintheTooth
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#15
In reply to #13

Re: torque required for tilting

08/30/2013 11:39 AM

OK so you have an issue with every european, and many Canadians who will answer the question " how much do you weigh ? " in kg.

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JBTardis
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#17
In reply to #15

Re: torque required for tilting

08/30/2013 4:24 PM

I am less annoyed by non-technical folks who misuse the term, but I do expect someone in a technical field to understand it and use it properly.

Reminds me of engineers (even those in the fluids industry) who will say something like "that hose is rated for 3000 pounds of pressure" instead of 3000 psi (pounds per square inch).

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K_Fry
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#16

Re: torque required for tilting

08/30/2013 2:11 PM

This question cannot be answered with the information given. Torque is expressed as a force x a lever arm distance. So without that distance (from the center of rotation to the center of gravity) how could the question possibly be answered?

Are you looking for the torque required at the motor or at the pivot point of the thing being tilted.

You can find the CG of an L shape graphically.

But maybe this is a trick question. During half of the rotation, positive torque is required. During the other half, the weight drives the motor, which you could call negative torque. The two will add up to 0.

If you imagine a perfect (utterly frictionless) flywheel, and give it a spin, if will spin forever, with no additional torque required. A perfect pendulum works the same way, even if your initial push sends the weight over the top.

Perhaps you can supply a drawing and your calculations so far.

Assuming this question was asked in a physics or engineering class, the distinction between weight and mass is an important one.

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Users who posted comments:

Anonymous Poster (2); dineshjegdeesh (2); JBTardis (2); K_Fry (1); LongintheTooth (3); lyn (1); nick name (1); PWSlack (3); Tornado (2)

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