• Support PF! Buy your school textbooks, materials and every day products Here!

Power required to rotate a load of specific weight

  • #1

Homework Statement


  • Weight - 200kg
  • RPM - 5
  • Diameter - 6m

Homework Equations


Moment of Inertia I = (weight/9.8).r^2
Torque t = I * (angular velocity / t)
Power = Torque * speed

The Attempt at a Solution


First I calculate
Moment of Inertia Using
I=(weight/9.8) * r^2 = 183 kg. m^2

Then I calculate Torque using
t=I∗(angularvelocity / t) = 96 Nm

where angular velocity is in rad/s.

Here is where I'm stuck if I take t as 1sec I get a Torque of 96 Nm.

If take t as 5 sec the I get a Torque of 19 Nm But my load is rotating at constant rpm of 5 so what would be the correct torque.

Once I get the torque I will find Power by P=Torque∗speed.
 
  • Like
Likes Delta2

Answers and Replies

  • #2
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
32,800
5,057
Unless there is friction, or some other opposing torque, no torque is required to maintain a steady rotation rate.
 
  • #3
Delta2
Homework Helper
Insights Author
Gold Member
2,425
684
Apparently the torque of the weight is the opposing torque. Thus we must provide a torque ##T_N## equal and opposite to the torque of weight ##T_B## so that the total torque ##T_N+T_B=0## is zero, so that the rotation rate is constant.

What is the torque of weight? Use the definition of Torque of a force, not from ##T=I\alpha## (to use the latter we will have to know the "partial" angular acceleration due to the torque of weight, which is unknown, though we know the total angular acceleration which is zero).

The Torque of weight ##T_B## will not be constant through time. It will depend on time t but not in the way you write in the OP.

Once you find ##T_B## use the relation for power ##P_B=T_B\omega## , where ##\omega## is the constant angular velocity , to find the power ##P_B## that the weight gives to its mass. The power of the torque ##T_N## will be equal and opposite that is ##-P_B##.
 
  • #4
Thanks for the esteemed answers. So to find the torque of weight Tb will i need to know the friction in the ball bearings on which the load rotates?
 
  • #5
Delta2
Homework Helper
Insights Author
Gold Member
2,425
684
Thanks for the esteemed answers. So to find the torque of weight Tb will i need to know the friction in the ball bearings on which the load rotates?
No , the Torque from the friction in the ball bearings is another torque, I assumed it to be zero for this exercise.

BUT tell me how exactly the weight rotates? Is the axis of rotation perpendicular to gravity or parallel to gravity?
 
  • #6
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
32,800
5,057
Apparently the torque of the weight is the opposing torque.
I suspect you are reading too much into the use of the word "weight".
As I read it, escape_velocity has a mass of 200kg rotating on axis, and probably the axis passes through the mass centre.
@escape_velocity , please clarify the exact set-up.
 
  • #7
Ok @Delta² @haruspex

So the actual setup is a cylinder (not a complete rod but a shell with thickness 5mm) with diameter 6 meter and weight 200kg and axis of rotation is passing through its centre. The axis of rotation is parallel to the earths surface and perpendicular to gravity.
 
  • #8
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
32,800
5,057
Ok @Delta² @haruspex

So the actual setup is a cylinder (not a complete rod but a shell with thickness 5mm) with diameter 6 meter and weight 200kg and axis of rotation is passing through its centre. The axis of rotation is parallel to the earths surface and perpendicular to gravity.
So without friction, why would any torque be required to maintain a constant rate of rotation?
 
  • #9
Delta2
Homework Helper
Insights Author
Gold Member
2,425
684
If that's the setup then the torque of weight is zero (assuming that the cylindrical shell has constant density and the axis of rotation passes through the c.o.m). There must be some friction force, either in the ball bearings or somewhere else (maybe in the point of contact between the cylindrical shell and the ground) in order for a torque to be needed to maintain a constant rate.
 
  • #10
Does that mean that once the cylinder is set into motion the only force that retards its rotation is the bearing friction. No it draws negligible power when its rotating at 5rpm its as though the motor is running unloaded. Is this the right deduction? Somehow it sounds strange.That a 200kg load is moving with negligible power!

But what would be the torque required to set this load into motion?
 
  • #11
CWatters
Science Advisor
Homework Helper
Gold Member
10,529
2,295
The relevant equation (if we ignore friction for the moment) is...

Torque = moment of inertia * angular acceleration

There is a similarly with the equation for linear motion..

Force = mass * acceleration

In both cases if the acceleration is zero (eg constant velocity or constant angular velocity) then the net force or torque is also zero.
 
  • #12
CWatters
Science Advisor
Homework Helper
Gold Member
10,529
2,295
To calculate the total torque needed to start it turning you need to know if there is any static friction as that can be higher than the running or kinetic friction. You also need to know how fast you want it to accelerate up to speed.
 
  • Like
Likes Delta2
  • #13
CWatters
Science Advisor
Homework Helper
Gold Member
10,529
2,295
If there is anything in the cylinder (eg its really a trommel used for gold mining) then that will have an effect.
 
  • #14
Delta2
Homework Helper
Insights Author
Gold Member
2,425
684
Does that mean that once the cylinder is set into motion the only force that retards its rotation is the bearing friction. No it draws negligible power when its rotating at 5rpm its as though the motor is running unloaded. Is this the right deduction? Somehow it sounds strange.That a 200kg load is moving with negligible power!
Yes this is because the torque of weight is zero , or almost zero since the axis of rotation passes through the center.
But what would be the torque required to set this load into motion?
if we ignore the static friction that @CWatters mentions then it depends on how fast you want it to catch the final angular velocity. If the final angular velocity is ##\omega## and you want it to catch this velocity in time ##t## then calculate the average angular acceleration as ##\alpha=\frac{\omega}{t}## and then the Torque will be ##T=I\alpha=mr^2\frac{\omega}{t}##
 

Related Threads on Power required to rotate a load of specific weight

Replies
2
Views
775
  • Last Post
Replies
5
Views
2K
Replies
2
Views
3K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
13
Views
3K
  • Last Post
Replies
8
Views
3K
Replies
6
Views
11K
Replies
20
Views
27K
  • Last Post
Replies
7
Views
3K
Top