# Homework Help: How to work out frictional torque

1. Dec 29, 2011

### lew123189

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

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"

Lewis

Last edited: Dec 29, 2011
2. Dec 29, 2011

### lew123189

Re: how to work out "frictional torque"

also, can you work out torque by:

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

3. Dec 29, 2011

### Staff: Mentor

Re: how to work out "frictional torque"

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. Jan 1, 2012

### michaelharmer

Re: how to work out "frictional torque"

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. Jan 4, 2012

### lew123189

Re: how to work out "frictional torque"

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. Jan 4, 2012

### lew123189

Re: how to work out "frictional torque"

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. Jan 4, 2012

### Staff: Mentor

Re: how to work out "frictional torque"

That would be the average velocity, not the velocity after 1.5 seconds.

8. Jan 4, 2012

### Staff: Mentor

Re: how to work out "frictional torque"

All forces acting on the hanging mass play a part. What forces act on it?

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. Jan 5, 2012

### lew123189

Re: how to work out "frictional torque"

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

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

but im not sure if this equation only applies when mass is falling free and hits the ground

10. Jan 5, 2012

### Staff: Mentor

Re: how to work out "frictional torque"

You found the average velocity. How does that relate to the final velocity, since the object started from rest?

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. Feb 4, 2012

### lew123189

Re: how to work out "frictional torque"

•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 with out accereration or the other way round.

12. Feb 4, 2012

### Staff: Mentor

Re: how to work out "frictional torque"

The first one will do just fine, since you have the time and the distance. Plug in the values and solve for the acceleration.
If you wanted to find the final velocity, you could use the fourth equation in your list. But you don't have to.

13. Feb 5, 2012

### lew123189

Re: how to work out "frictional torque"

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. Feb 5, 2012

### Staff: Mentor

Re: how to work out "frictional torque"

Yes, you need to solve for a.
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. Feb 5, 2012

### lew123189

Re: how to work out "frictional torque"

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. Feb 6, 2012

### lew123189

Re: how to work out "frictional torque"

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 =
=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 in to it.

thanks a lot

17. Feb 6, 2012

### Staff: Mentor

Re: how to work out "frictional torque"

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. Oct 25, 2012

### saucysaunders

Re: how to work out "frictional torque"

Hi, can I continue with this question?

Using the last post I find:
Angular acceleration = linear accel / radius

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 = 0.942Nm

So, 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: Oct 25, 2012
19. Oct 25, 2012

### Staff: Mentor

Re: how to work out "frictional torque"

Good.

Use this to find the tension in the rope. Once you find the rope tension, then you can find the torque it produces.

Redo this calculation.

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. Nov 8, 2012

### saucysaunders

Re: how to work out "frictional torque"

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