# Calculating torque to move an object

Can somebody please tell me how I calculate how much torque is required to turn a wheel to move an object weighing 80Kgs

 ebaines Posts: 10,583, Reputation: 5794 Expert #11 Nov 18, 2008, 05:54 AM

Quote:
 Originally Posted by gmlittle_01 OK so if a = 1m/s²
Just plug 1 m/s^2 into the formula I gave you to determine the torque required from your motor to accelerate at 1 m/s^2:

Torque = 80 Kg * 0.15m * a
= 12 Newton-meters

The velocity would be V = 1m/s^2 * t, so it would take 5 seconds to reach 5 m/s.

Visharad - your approach assumes that the problem here is to spin a wheel (i.e. torque used to overcoe the inertia of the wheel). But this problem is different - it's to get a scooter moving forward, and I think you can assume that the inertia of the wheel is insignificant.
 visharad Posts: 47, Reputation: 15 Junior Member #12 Nov 18, 2008, 07:54 AM

Yes, I assumed that the question is to find torque to spin the wheel. That is what the question appeared to me.
It will be good if the gmlittle_01 posts the complete question in a single post. Then it will be quite clear what he really wants to know.
 gmlittle_01 Posts: 7, Reputation: 1 New Member #13 Nov 20, 2008, 06:50 PM
Hi Guys, thanks for all you help, hopefully the following will make things clear:

I would like to find out what torque is required to turn the wheels (Diameter = 0.15m) of a scooter which is being riden along the ground carrying a person that weighs 80KG.

Is it also possible to calculate what the torque is from a standing start i.e to start turning the wheels, and then what the torque is required to keep the scooter traveling at 5m/s.

It would also be useful to know how the torque changes when the scooter is travelling up hill keeping at a constant speed of 5m/s. You will probably need to know what the steepness of the hill is, but maybe this can be plotted on a graph which shows how the torque increases versus the steepness.
 visharad Posts: 47, Reputation: 15 Junior Member #14 Nov 20, 2008, 09:09 PM

Are you trying to find the following?
1. There is a scooter, whose wheels are of diameter 0.15 m each. The only person seated on the scooter is the driver. His weight is 80 kg. The wheels are not moving and the scooter is at rest .
What torque is needed just to make the wheels start rotating?

2. Now the wheels have just started rotating. What torque is needed to increase the speed of the scooter to 5 m/s?

3. The scooter is moving at 5 m/s. What torque is needed to maintain this 5 m/s speed?

4. The scooter is moving horizontally at 5 m/s. Now the scooter has to go on a hill of a given angle of inclination. But the speed should remain at 5 m/s. What torque is needed to bring the scooter from the horizontal to the hill?

5. How the torque in (4) depends on the angle of inclination of the hill?

Is my understanding of your questions correct?
 gmlittle_01 Posts: 7, Reputation: 1 New Member #15 Nov 20, 2008, 09:36 PM

Yes all correct
 visharad Posts: 47, Reputation: 15 Junior Member #16 Nov 20, 2008, 09:50 PM

Radius R = 0.15 m/2 = 0.075 m
I am calculating torque on a single wheel about the axis of this wheels.

1)Let u = coefficient of static friction between the wheels and the road,
M = mass of the scotter + driver on it
Sum of normal forces on the two wheels by the road = weight = Mg
Normal force on one wheel = Mg/2
Maximum static friction on the wheel = u*Mg/2
Force on that wheel required to make the wheel turn = u*Mg/2
Torque on the wheel = u*(Mg/2)*R = (u*Mg*R)/2

This is the torque required to make one wheel start turning.
I have assumed that there is symmetry in the sense that normal forces and friction forces on the wheels are equal.
 visharad Posts: 47, Reputation: 15 Junior Member #17 Nov 20, 2008, 09:56 PM
2)Initial speed vo = 0
Final speed v = 5 m/s
Initial angular speed wo = vo/R = 0
Final angular speed w = v/R
Let time taken to change the speed = t
Angular acceleration alpha = (w - wo)/t = w/t = v/(Rt)
Let moment of inertia of the wheel = I

Then torque = I * alpha
= I * v/(Rt)
For a given value of v, you can find torque as a function of t according to the above equation.
The value of I depends on the shape of the wheel and the mass distribution in it.
Note: You will need mass to find the value of I. You should use mass of only the wheel here.
 sahil2493 Posts: 1, Reputation: 10 Junior Member #18 Mar 18, 2012, 08:17 AM
can the motor shown in the link below move a person at a speed of 20km/h taking acceleration to be 6m/(s^2) provided the dia of the wheel is 15cm...???
consider that 4 of these motors share the load equally.
http://robokits.co.in/shop/index.php?main_page=product_info&cPath=2&products_ id=328

if not can this motor satisfy the required conditions with a little less speed
http://robokits.co.in/shop/index.php?main_page=product_info&cPath=2&products_ id=212
 ebaines Posts: 10,583, Reputation: 5794 Expert #19 Mar 19, 2012, 07:34 AM

The torque that the motor(s) need to provide is equal to:

$
\tau = mar
$

where m = mass of the man (you don't provide a number so let's assume 100 Kg), a = acceleration of theman = 6m/s^2, and r =radius of the gear that is driving the conveyor belt = 0.15/2 m.
So:

$
\tau= 100 Kg \ \times \ 6 \frac m {s^2} \ \times \ 0.075m \ = \ 45\ N-m
$
 karan_utd Posts: 2, Reputation: 1 New Member #20 Apr 10, 2012, 10:45 AM
any idea of the torque recquired to rotate the gear section present on the handle bar of a scooter

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