How to work out frictional torque

In summary, to work out the frictional torque in a system with a suspended mass and flywheel, one must first calculate the linear and angular acceleration of the mass and flywheel, respectively. Then, use the equations ƩF = m*a and ƩTorque = I*alpha to determine the net forces and torques acting on the system. The frictional torque can be found by subtracting the torque due to the hanging mass from the net torque. Additionally, the tension in the rope can be calculated by considering all forces acting on the hanging mass, including gravity. The maximum velocity of the mass can be found by equating its potential and kinetic energies.
  • #1
lew123189
13
0
how to work out "frictional torque"

A Mass of 0.5 kg is suspended from a flywheel. If the mass is released from rest and falls a distance of 0.5 m in 1.5 s, calculate
the tension on the rope
the frictional torque, resisting motion

Mass of flywheel = 3 kg
radius of wheel = 300mm
radius of gyration = 121mm

so this is what i have worked out so far

velocity = d/t = 0.5/1.5 = 0.33
angular acceleration = (max mass velocity - initial velocity)/time = 0.33- 0 / 1.5 = 0.22

now i think tension = mass X (gravity X acceleration)
however i am not sure what acceleration they mean is it the angular or linear acceleration.

Torque = moment of inertia X angular acceleration
= (0.5X0.212squared)X0.22
= 0.022472 X 0.22 = 4.94X10 to the power of -3
so what i need to know is how to find the "frictional torque" and the "resisting motion"

thanks in advance


Lewis
 
Last edited:
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  • #2


also, can you work out torque by:

Mass X radius of pulley

(0.5X9.81(gravity))X0.3 = 1.4715N
 
  • #3


You need to apply Newton's 2nd law to the flywheel and to the hanging mass:

ƩTorque = I*alpha (for the flywheel)

ƩF = m*a (for the hanging mass)

Start by identifying the forces acting on each.

Note:
ƩTorque means net torque; there are two torques acting here.

Similarly, ƩF means net force.
 
  • #4


angular acceleration = (max mass velocity - initial velocity)/time = 0.33- 0 / 1.5 = 0.22

This is not angular acceleration, This is the calculation for linear acceleration.

a) The linear acceleration of the mass.

v = s / t
a = v / t

b) The angular acceleration of the wheel

Omega = v / r

Knowning that v = Linear Velocity / r = radius of flywheel
This provides angular speed (rad s^-1)

Alpha = Omega / t

c) The tension in the rope.

F = mg - ma

Is this equation correct?

d) The frictional torque, resisting motion

T = m x k^2 x aplha. This is accelerating torque

Just struggling to understand how to calculate the Total Torque required.

Is Tnet F = ma, so mass of flywheel by a

Then frictional torque is

Tf = Tnet - Ta

Does this seem correct?
 
  • #5


Ok so this is were I am so far after taking your advise

Linear acceleration = mass max velocity – initial velocity/ time
= 0.33-0/1.5 = 0.22 ms -2

Angular acceleration = max wheel speed – initial wheel speed / time
Max wheel speed = velocity / radius = 0.33/0.3 = 1.1 rad s -1

So angular acceleration = 1.1 – 0 / 1.5 = 0.73 rad s -2

Now is net torque = moment of inertia X angular acceleration

Moment of inertia = MK 2
= Mass of wheel X radius of gyration squared
= 3X0.212 squared
=0.134 KG m squared
Net torque = 0.134 X0.73
= 0.9842 NM

Net force = MXA
= 0.5 X 0.219
= 0.1095

So would frictional torque be net torque – force of hanging object
= 0.9842-0.1095 = 0.8747 NM
Frictional torque = 0.8747 NM ?
 
  • #6


could the tension in the rope simply be mass X acceleration = 5 X 0.219 = 1.095

or does gravity have to play a part?

would this make more sense?
tension = mass X (gravity X acceleration)
= 5 X(9.81 X 0.219) = 10.74 ?
 
  • #7


lew123189 said:
so this is what i have worked out so far

velocity = d/t = 0.5/1.5 = 0.33
That would be the average velocity, not the velocity after 1.5 seconds.
 
  • #8


lew123189 said:
could the tension in the rope simply be mass X acceleration = 5 X 0.219 = 1.095

or does gravity have to play a part?
All forces acting on the hanging mass play a part. What forces act on it?

would this make more sense?
tension = mass X (gravity X acceleration)
= 5 X(9.81 X 0.219) = 10.74 ?
No.

As I suggested before, apply:

ƩF = m*a (for the hanging mass)

ƩTorque = I*alpha (for the flywheel)

Start by figuring out the correct linear acceleration of the mass.
 
  • #9


I see that makes sense. I am struggling to figure out maximum velocity.
How about this

VMAX

Potential energy of mass = MGH
=0.5X9.81X0.5
= 2.45 J

Kinetic energy at bottom = potential energy at top

Thus ½ MN 2 = 2.45J

Were V is velocity immediately before bottom
Therefore ½ X 0.5 X V 2 = 2.45

V = square root of 2.45 X2 /0.5

Velocity immediately before bottom = 3.13 ms -1

Maximum velocity of mass = 3.13 ms -1

Therefore maximum velocity of pully = Vmax / radius = 3.13 / 0.3 = 10.43 rad S –1

= 10.43 rad S-1


but I am not sure if this equation only applies when mass is falling free and hits the ground
 
  • #10


lew123189 said:
I see that makes sense. I am struggling to figure out maximum velocity.
You found the average velocity. How does that relate to the final velocity, since the object started from rest?

How about this

VMAX

Potential energy of mass = MGH
=0.5X9.81X0.5
= 2.45 J

Kinetic energy at bottom = potential energy at top
That won't work because you've neglected the kinetic energy of the flywheel. (Total mechanical energy is conserved.)

How about forgetting about the velocity and solving for the acceleration directly? What kinematic formula relates distance, time, and acceleration?
 
  • #11


•d = vo • t + 0.5 • a • t2
•vf = vo + a • t
•vf2 = vo 2 + 2 • a • d
•d = (vo + vf)/ 2 • t
where

•d = displacement
•t = time
•a = acceleration
•vo = original or initial velocity
•vf = final velocity

these are the only kinematic equations i could find. seems hard to find the final velocity without accereration or the other way round.
 
  • #12


lew123189 said:
these are the only kinematic equations i could find.
The first one will do just fine, since you have the time and the distance. Plug in the values and solve for the acceleration.
seems hard to find the final velocity without accereration or the other way round.
If you wanted to find the final velocity, you could use the fourth equation in your list. But you don't have to.
 
  • #13


I see I guess I have to make a the subject?
•d = vo • t + 0.5 • a • t2
I'm not very good at this
D-1/2xaxt2=v1xt
1/2xaxt2=[v1xt2]+d
So
1/2a = (v1xt2)+d/t2
How does that look? Also by t2 are they referring to time again?
Cheers
 
  • #14


lew123189 said:
I see I guess I have to make a the subject?
Yes, you need to solve for a.
•d = vo • t + 0.5 • a • t2
Better to write this as:
d = v0t + 0.5at2

I would plug in the values and then solving for a will be easy.

What are the values of d, v0, and t?
 
  • #15


Apologise in advance I'm not good at these
D=vot+0.5at2
=0.5=0x1.5+0.5xax1.5(2)
=0.5=0.5xax1.5(2)
=ax1.5(2)=0.5/0.5
=ax1.5(2)=1
=a=1/1.5(2)
=a=1/2.25
=a=0.44
Cheers
 
  • #16


OK so

Linear accereration = V/T
=0.33/1.5 = 0.22

Angular acceleration =
2 (D-VoT)/T(2) = A
A = 2(0.5-0)/1.5(2) = 0.44
Angular acceleration = 0.44

VF= final velocity
VF=Vi+AT
VF=0+0.44X1.5
VF=0.66

Moment of inertia =
=MXradius of gyration (2)
=0.5X0.212(2) = 0.022472

Net torque of wheel = IXalpha
= 0.22472X0.44
=0.0988 N

Net force of hanging mass = MxA
= 0.5X0.22
=0.11 N

How does this all look?

still not sure how "frictional torque" and the "resisting motion" goes into it.

thanks a lot
 
  • #17


lew123189 said:
Apologise in advance I'm not good at these
D=vot+0.5at2
=0.5=0x1.5+0.5xax1.5(2)
=0.5=0.5xax1.5(2)
=ax1.5(2)=0.5/0.5
=ax1.5(2)=1
=a=1/1.5(2)
=a=1/2.25
=a=0.44
Cheers
OK, this is the linear acceleration of the falling mass (in m/s^2).

Use this to:
Figure out the angular acceleration of the flywheel.
Figure out the tension in the rope.
 
  • #18


Hi, can I continue with this question?

Using the last post I find:
Angular acceleration = linear accel / radius
=0.44/0.3 = 1.481 rad/s^2

Then Doc as you said previously find net force and torque:

Net Force = ma = 0.5*0.44 = 0.22N

Net Torque = I*alpha = mk^2 *alpha = 0.636 * 1.481 = 0.942NmSo, providing this is correct, how do I then find the frictional torque? or which one is which etc?Would I just need to add the two together?
 
Last edited:
  • #19


saucysaunders said:
Hi, can I continue with this question?

Using the last post I find:
Angular acceleration = linear accel / radius
=0.44/0.3 = 1.481 rad/s^2
Good.

Then Doc as you said previously find net force and torque:

Net Force = ma = 0.5*0.44 = 0.22N
Use this to find the tension in the rope. Once you find the rope tension, then you can find the torque it produces.

Net Torque = I*alpha = mk^2 *alpha = 0.636 * 1.481 = 0.942Nm
Redo this calculation.

So, providing this is correct, how do I then find the frictional torque? or which one is which etc?
You'll solve for the frictional torque using Newton's 2nd law for rotation:
ƩT = I*alpha

You have two torques acting: Trope & Tfriction (acting in opposite directions).
So ƩT = Trope - Tfriction = I*alpha
 
  • #20


Thanks for the reply!
So,

Tension in rope F = ma + mg = (0.5*0.44)+(0.5*9.81) = 5.127N

The torque for the rope is therefore Trope = 5.127*0.212 = 1.087Nm


Then, from my mistake at squaring earlier (thanks for pointing out):

Net Torque = I*alpha = mk^2 *alpha = 0.135 * 1.481 = 0.199935Nm = 0.2Nm


So therefore TFriction = 1.087 - 0.2 = 0.887Nm


Does that sound right? I was expecting a bit less friction I suppose...But maybe I'm just doubting my own competence now. haha
 
  • #21


saucysaunders said:
So,

Tension in rope F = ma + mg = (0.5*0.44)+(0.5*9.81) = 5.127N
Careful with signs:
ƩF = ma
F - mg = ma

But the acceleration is downward, thus negative. (Using the up = + convention.)
 
  • #22


I understand acceleration is downwards, but using the neg figure would give tension as -5.127N

Can you have a negative tension? Should I therefore be using this negative answer to work out the Rope torque?
 
  • #23


saucysaunders said:
I understand acceleration is downwards, but using the neg figure would give tension as -5.127N
That's not correct. Show how you got that. (Only the 'a' is negative.)
 
  • #24


If your using the Newton's second law for rotational motion
τ=αI
You have to make sure the units are correct.
Torque is not Nm it's J/rad.
1 Nm is 2π J/rad
Moment of Inertia stays as kg.m2

Going back to basic units
[(kg.m^2)/(s^2)]/(rad)=(kg.m^2)(rad/s^2)
 
  • #25


motoxYogi said:
If your using the Newton's second law for rotational motion
τ=αI
You have to make sure the units are correct.
Torque is not Nm it's J/rad.
1 Nm is 2π J/rad
Moment of Inertia stays as kg.m2

Going back to basic units
[(kg.m^2)/(s^2)]/(rad)=(kg.m^2)(rad/s^2)
Standard units for torque are N-m. (You can use J/rad, but that isn't helpful here.)
 
  • #26


I was confusing myself, by ignoring the fact that the wheel turning actually makes the tension in the rope less. It's all coming together now I think...

Tension in the rope Ft = (0.5* -0.44) + (0.5*9.81) = 4.685N
Torque in the rope = 4.685 * 0.212 = 0.993Nm

So Frictional Torque = 0.993 - 0.2 = 0.793Nm (Lower than previous answer as I said before)


If I then convert to J/rad as moto explains:

0.793Nm * 2π = 4.98 J/rad
 
  • #27


Doc Al said:
Standard units for torque are N-m. (You can use J/rad, but that isn't helpful here.)

It is if you're using angular acceleration rad/s2 not m/s2
 
  • #28


motoxYogi said:
It is if you're using angular acceleration rad/s2 not m/s2
Torque = I*alpha, where torque units are N-m and alpha units are rad/s2.
 
  • #29


saucysaunders said:
I was confusing myself, by ignoring the fact that the wheel turning actually makes the tension in the rope less. It's all coming together now I think...

Tension in the rope Ft = (0.5* -0.44) + (0.5*9.81) = 4.685N
Torque in the rope = 4.685 * 0.212 = 0.993Nm

So Frictional Torque = 0.993 - 0.2 = 0.793Nm (Lower than previous answer as I said before)
Good.
If I then convert to J/rad as moto explains:

0.793Nm * 2π = 4.98 J/rad
Don't do that!
 
  • #30


Your wish is my command oh great one :P

Thanks for the help!
 
  • #31


Doc Al said:
Torque = I*alpha, where torque units are N-m and alpha units are rad/s2.

What you are saying is that

N.m = kg.m2(rad/s2)
or(kg.m/s2)(m) = kg.m2(rad/s2)

or am I missing something?
 
  • #32


motoxYogi said:
What you are saying is that

N.m = kg.m2(rad/s2)
or(kg.m/s2)(m) = kg.m2(rad/s2)

or am I missing something?
Yes. Dimensionally, radians have no significance.
 
  • #33


Doc Al said:
Yes. Dimensionally, radians have no significance.

I apologize for wasting everyones time then. Sorry
 
  • #34


saucysaunders said:
I was confusing myself, by ignoring the fact that the wheel turning actually makes the tension in the rope less. It's all coming together now I think...

Tension in the rope Ft = (0.5* -0.44) + (0.5*9.81) = 4.685N
Torque in the rope = 4.685 * 0.212 = 0.993Nm

So Frictional Torque = 0.993 - 0.2 = 0.793Nm (Lower than previous answer as I said before)


New to the forum thing! Working through the problem I am following it as far as the tension in the rope, but am a little confused by the next line - torque in the rope, which is obviously the tension * the radius of gyration, why are we multiplying these two together? Any help appreciated.
 
Last edited by a moderator:
  • #35


should it not be the radius of the pulley, 300mm? that we are multiplying by 4.685?
 
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