## how to calculate required hp for machine

one rubber coated roll weight 1000 kg. dia. 300 mm & face length of 6360 mm. bothe end shaft dia. 65 rotating in antifrictional bearings.

i want to rotate the roll at 100 rpm within 60 seconds.

so please help me out to find the HP required of geared motor of 1400 rpm input of motor & gearbox ratio is approx 25:1

its very urgent
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 if you need an answer so badly why don't you explain your problem better? this is just lazy: "bothe end shaft dia. 65" have respect for people trying to help you. tell us what you have done to solve the problem. what approaches have you made so far? did you at least look it up in wikipedia? was that not enough?
 "i want to rotate the roll at 100 rpm within 60 seconds." you realize RPM stands for rotations per minute, right?

## how to calculate required hp for machine

1 hp = 745.7 watts

1 watt = 1 W = 1 J/s = 1 N·m/s = 1 kg·m2/s3
 but how can you calculate 1 hp ?
 We are going to need to know more about the makeup of the roll. Do you have a drawing with dimensions? We need to be able to calculate a mass moment of inertia for the roll.

 Quote by OldEngr63 We are going to need to know more about the makeup of the roll. Do you have a drawing with dimensions? We need to be able to calculate a mass moment of inertia for the roll.
Can we not approximate a solid cylinder of 300 mm dia and 6,360 mm in length? And 1,000 kg. I suspect he doesn't need the answer correct to 2 d.p. (unless it is for homework).

I feel we ought to do something for the OP who may have been frightened off of further posting by some of the 'forthright' replies.

Attached Files
 DRAG ROLL.pdf (75.2 KB, 30 views)

 Quote by harshadeng i want to rotate the roll at 100 rpm within 60 seconds. so please help me out to find the HP required of geared motor of 1400 rpm input of motor & gearbox ratio is approx 25:1 its very urgent
Do you have a variable speed motor?

If you have 1400 RPM, and you reduce it by a factor of 25:1 (through the gearbox) the resulting RPM will be 56. You can't get 100 RPM with this setup unless you can vary the input rpm. To get 100 RPM, you need a 2500 RPM motor hooked up to the 25:1 gearbox. Or, you could use a 1000 RPM motor hooked up to a 10:1 gearbox.

Your motor's RPM is decreased by reduction ratio of the gearbox. Your motor's torque is increased by the same ratio.
 thank u but sorry my mistake was there motor rom is 2500 i had attached a drawing of the roll earliar for referance & to calculate MOI
 Bandit127, go right ahead if you want to mislead the man. If yoou look at the drawing that he posted, treating the roll as a solid cylinder will produce a grossly erroneous MMOI value. That is why it is necessary to ask for information.

 Quote by OldEngr63 Bandit127, go right ahead if you want to mislead the man. If yoou look at the drawing that he posted, treating the roll as a solid cylinder will produce a grossly erroneous MMOI value. That is why it is necessary to ask for information.
Yes, I take your point - a solid cylinder would have underestimated the power required. Perhaps even by a gross amount.

In fairness, my post was before we had the drawing though.

I did have a look at the maths but it was too far from my field for me to get a good hold of it, so I will leave the calculations to the experts.

Good to see you are still in the thread though harshadeng. While I am sure they weren't meant that way, some of the initial posts read as unnecessarily critical to me and could have put some people off.
 The moment of inertia for a solid cylinder is $I=\frac{1}{2}mr^{2}$, while the moment of inertia of a hollow cylinder is $I=mr^{2}$. The energy of a rotating cylinder is $E=\frac{1}{2}I\omega^2$. Dividing E by the acceleration time τ gives the power required (in watts). So the power required is about $$E=\frac{1}{2\tau}m r^2 \omega^2$$ Using τ=60, ω=2π100/60, r=0.15, and m=1000, The power to required accelerate to 100 RPM in 60 seconds is about 21 watts (constant power solution). If the acceleration rate is constant, the torque is $T=I\frac{d\omega}{dt}=m r^2\frac{d\omega}{dt}=3.9 \text{ Newton-meters}$ and the maximum power is $P=T\omega= 3.9\omega_{max}=41 \space watts$
 Would you like to comment on the realism of your results, Bob S?

 Quote by OldEngr63 Would you like to comment on the realism of your results, Bob S?
What means realism? Isn't this the moment of inertia of a solid cylinder?
$$I=\int_{V}\rho r^2 dV=2\pi\int_{0}^{r}\rho \ell r^3 dr=\frac{1}{2}Mr^2$$
 Realism means that you really believe that 41 watts will do this job. I don't. Look at your overall calculation and tell us that the whole scenario is realistic, if you believe it is.

 Quote by OldEngr63 Realism means that you really believe that 41 watts will do this job. I don't. Look at your overall calculation and tell us that the whole scenario is realistic, if you believe it is.
What do you think the total rotational energy of the cylinder is, in Joules?