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- Thread starter siva surya
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billy_joule

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It may be that:

T = Fr where the force is due to the 10kg mass and r is the radius of the axle?

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billy_joule

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in this

1.how much torque is reuired to rotate axle of diameter 10 cm

2.and on the other end wheel is attached of diameter 100 cm and hang with 10 kg of weight

3.then in how many kg should be axle and wheel metal ?

1. It depends on how fast you want the system to accelerate.

T = Iα

Where T is torque, I is moment of inertia and α is angular acceleration.

The moment of inertia depends on geometry and mass:

https://en.wikipedia.org/wiki/Moment_of_inertia

The above only considers the torque required to accelerate the axle & pulley, additional torque will be required to accelerate the hanging mass upwards.

3. The mass of the axle & wheel will be the volume multiplied by density. The required geometry will depend on the forces applied, in other words it needs to be of a certain size (and mass) so it doesn't break.

Is this homework? Or a real project? Maybe you are trying to design a hand winch?

https://www.google.co.nz/search?q=h...Cc#q=hand+winch&tbm=isch&tbs=isz:lt,islt:svga

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i like to design no energy machine

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billy_joule

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A machine, by definition, uses energy.

https://en.wikipedia.org/wiki/Machine

https://en.wikipedia.org/wiki/Machine

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ya like to use gravity as energy

- #8

CWatters

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1.how much torque is reuired to rotate axle of diameter 10 cm

2.and on the other end wheel is attached of diameter 100 cm and hang with 10 kg of weight

The torque must be at least..

= (10Kg * 9.81) * 100/10

= 981 Nm

That torque will stop the weight falling. If you want to raise the weight then (as billy said) you need to tell us how fast you want it to accelerate and the moment of inertia of the pulleys etc. Friction?

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if the 10 kg mass falls as free fall and create impact force means whats the impact force on there

in which mass 10 kg

height 20 cm

whats newton there

in which mass 10 kg

height 20 cm

whats newton there

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- #10

CWatters

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If you assume it hits something like modelling clay and has constant deceleration you can estimate using..

V

where

U

V

a = acceleration

s = distance (eg depth of impact crater)

So you could measure the depth of the impact crater and plug in the numbers to estimate the deceleration "a".

Then use F = ma to estimate the implied impact force.

The results will be very approximate. In the real world it's better to attach an accelerometer to the mass and measure "a".

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how much torque is required to pull up to weight 11 kg from 100 cm through wheel and axle?

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If you assume it hits something like modelling clay and has constant deceleration you can estimate using..

V^{2}= U^{2}+2as

where

U^{2}= velocity at impact

V^{2}= velocity after impact, probably zero

a = acceleration

s = distance (eg depth of impact crater)

So you could measure the depth of the impact crater and plug in the numbers to estimate the deceleration "a".

Then use F = ma to estimate the implied impact force.

The results will be very approximate. In the real world it's better to attach an accelerometer to the mass and measure "a".

how much torque is required to pull up to weight 11 kg from 100 cm through wheel and axle?

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110 Nm

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CWatters

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how much torque is required to pull up to weight 11 kg from 100 cm through wheel and axle?

In post #8 I showed you how to calculate the torque for 10kg...

The torque must be at least..

= (10Kg * 9.81) * 100/10

= 981 Nm

For 11kg the sum becomes..

= (11Kg * 9.81) * 100/10

= 1079 Nm

or has your system changed?

- #15

CWatters

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PS: It's hard to understand your English.

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