Kenetics torque tension acceleration question

[electro82]

Homework Statement

i am not looking for the answers here just someone to steer me back on track as i feel i am moving away from the correct path.

i have answered some of the problem and i am a little unsure if i have tackled it in the correct way.

a mass of 0.5kg is suspended from a flywheel which has a mass of 3kg, a radius of 300mm and a radius of gyration of 212mm.
the mass is released from rest and falls 0.5m in 1.5 seconds i need to calculate:

A.the linear acceleration of the mass.

B. the angular acceleration of the flywheel.

C. the tension in the rope. this is where i start to get a bit lost

D. the frictional torque, resisting motion.
as i said i am not looking for someone to give me the answers i am looking for a little guidence, any help would be appriciated.

Homework Equations

a=2s/t^2
alpha =a/r
T=mg-ma
T(friction)=T(net)-T(acceleration).

The Attempt at a Solution

A. I have 0.44444ms^-2 i used a=2s/t^2 is this the correct method?

B. i have 1.481481481 rads^-2 i used alpha =a/r again is this the correct method?

C. i have 4.682778N i used T=mg-ma is this correct?

D. is this not the same as the tension as this is the force opossing mg to reduce a from 9.81ms^-2 to 0.444444ms^-2?

so i have 4.682778N of force have calculated the frictional torque as T(acceleration)=I x alpha.
how do i calculate the net resisting torque as i think that the frictional torque is
T(friction)=T(net)-T(acceleration).

any help to put me back on track would be greatly appriciated.

 

[tiny-tim]

welcome to pf!

hi electro82! welcome to pf!

(have an alpha: α and a tau: τ and try using the X² icon just above the Reply box )

A. I have 0.44444ms^-2 i used a=2s/t^2 is this the correct method?

B. i have 1.481481481 rads^-2 i used alpha =a/r again is this the correct method?

C. i have 4.682778N i used T=mg-ma is this correct?

D. is this not the same as the tension as this is the force opossing mg to reduce a from 9.81ms^-2 to 0.444444ms^-2?

so i have 4.682778N of force have calculated the frictional torque as T(acceleration)=I x alpha.
how do i calculate the net resisting torque as i think that the frictional torque is
T(friction)=T(net)-T(acceleration).

any help to put me back on track would be greatly appriciated.

your methods for A to C are correct (I haven't checked the figures)

i haven't quite followed your D …

the frictional torque is (presumably) from the axle, and the net torque is the torque from the rope minus the frictional torque, with τ_net = Iα

 

[electro82]

for D
if the mass falls @ 9.81ms^-2 there is no resisting force.
the mass actually falls @ 0.4444ms^-2

so using I x alpha i calculate the acclerating torque this is the torque required to accelerate the flywheel?

but this is where i start to get confused i know the linear force that reduces the acceleration from 9.81 to 0.4444 but how do i convert this force into an angular figure i.e. the resisting torque?

am i right that I x alpha is the acclerating torque i.e. the torque required before the flywheel can accelerate? if so the frictional torque would be the total torque opossing the gravitational force on the mass minus this accelration torque.

as i said this is where i get a little confused.

also why is the tension in the rope not the gravitaional force of the mass plus the resisting force on the rope reducing the acceleration?

thanks for your replies

 

[tiny-tim]

hi electro82!

am i right that I x alpha is the acclerating torque …

(what happened to that α i gave you? )

Iα is equal to the net torque (just as ma is equal to the net force in the linear case)

i.e. the torque required before the flywheel can accelerate?

"before" ?

i know the linear force that reduces the acceleration from 9.81 to 0.4444 but how do i convert this force into an angular figure i.e. the resisting torque?

you should know this …

torque = force times (perpendicular) distance, r x F

if so the frictional torque would be the total torque opossing the gravitational force on the mass minus this accelration torque.

as i said this is where i get a little confused.

also why is the tension in the rope not the gravitaional force of the mass plus the resisting force on the rope reducing the acceleration?

please stop talking, or thinking, about "resisting force"

there's frictional force, tension, gravitational force, and so on, and together they make up net force, but there's no such thing as resisting force …

i'm sure that you're confusing yourself by talking about it (and you're certainly confusing me)

always draw a diagram, identify the forces (or torques), add them to get the net force (or torque), and then equate it to ma (or Iα) ​

 

[electro82]

the torque as a result of the Mass would be 4.905N x 0.3m =1.4715Nm
the frictional torque (is this the correct term?) is 4.682778N x 0.3m = 1.4048334Nm
so the net torque would be = 0.2222222N x 0.3m = 0.0666666Nm

my answer to the D therefore would be that the frictional torque (resisting motion) is 0.0666666Nm?

from a diagram 4.905N down (mg) 4.682778N up (mg-ma) the net force 0.2222222N (ma)

this was all calculated using the linear figures, is there a way of calculating the torque using the angular data?

thank you for your patience i think that you could be correct when you say i have confused myself!

 

[tiny-tim]

A.the linear acceleration of the mass.

B. the angular acceleration of the flywheel.

C. the tension in the rope. this is where i start to get a bit lost

D. the frictional torque, resisting motion.

…
A. I have 0.44444ms^-2 i used a=2s/t^2 is this the correct method?

B. i have 1.481481481 rads^-2 i used alpha =a/r again is this the correct method?

C. i have 4.682778N i used T=mg-ma is this correct?

the torque as a result of the Mass would be 4.905N x 0.3m =1.4715Nm

sorry, what are these figures?

the frictional torque (is this the correct term?) is 4.682778N x 0.3m = 1.4048334Nm

no, that is the torque of the tension …

the frictional torque (D., in the question) is the torque at (presumably) the axle, and is what you have to find by subtracting the tension torque from Iα

 

[electro82]

sorry for the silence i have been away

but hopefullyi have got this one now

right so to calculate the acelerational torque i use I x Alpha
I=mk^2 = 3 x .212^2 x 1.481481481 = 0.199751046Nm

this is the torque required to accelerate the flywheel to the 1.481481481rads^-2 ?

total torque = 4.68278 x 0.3 = 1.2804834Nm (tensional torque)

therefore frictional torque = 1.280483Nm - 0.199751046Nm = 1.080732354Nm

please tell me i am now on the right track

 

[tiny-tim]

hi electro82!

you Iα term looks ok

i don't follow where your total torque comes from

 

[electro82]

thank you for being so patient i am sure i have made this much more complicated than it really is

so Iα calculates the torque required to accelerate the flywheel to α ?

by total torque i mean the force of the mass (mg) x r = 0.5kg x 9.81ms^-2 x 0.3m =
1.4715Nm that would be the torque on the pivot due to the mass?

i think i may need to go back to the beginning on this one as i said before i have confused myself i think?

could you break down the steps to find D with a brief explanation of what each step is telling us, as i think i am not understanding some of the principles to this subject?

again i can not thank you enough for your help and i can only appologies for my constant questions.

 

[tiny-tim]

hi electro82!

a mass of 0.5kg is suspended from a flywheel which has a mass of 3kg, a radius of 300mm and a radius of gyration of 212mm.
the mass is released from rest and falls 0.5m in 1.5 seconds

so Iα calculates the torque required to accelerate the flywheel to α ?

by total torque i mean the force of the mass (mg) x r = 0.5kg x 9.81ms^-2 x 0.3m =
1.4715Nm that would be the torque on the pivot due to the mass?

ah, no, the torque τ is not mgr, because the tension T is not mg

from the given data, you can find a and α

then you need to do F = ma for the mass, and τ = Iα for the flywheel

(they are connected by τ = rT and a = rα)

 

[electro82]

Tension = mg-ma = 4.68278N

Torque = (mg-ma) x r = 1.404834Nm

Iα = 0.199751046Nm this is acceleration Torque (required to create α)?

so fricitonal Torque=Torque-acceleration Torque (Iα)

i calculate 1.40483Nm-0.199751046Nm= 1.205078954Nm

frictional torque = 1.205078954Nm

i hope this is correct

so can I simply covert the linear mg, ma & mg-ma into torque by multiplying the force by r

mg= 4.905n so mgr=1.4715Nm
mg-ma= 4.68278N so (mg-ma)r = 1.404834Nm
ma= 0.22222 so mar= 0.066666Nm

if this is all correct i owe you a big thank you for your persistance.biggrin: if not can you see where i am going wrong or what i have misunderstood?

 

[tiny-tim]

yes, that's all fine … congratulations!

btw, i checked by using a system you're probably not allowed to use, that avoids finding the tension …

we pretend that the flywheel is a non-rotating sliding block of "rolling mass" I/r² = Mr_gyr²/r² = 1.5 kg (do you notice how they've chosen all the figures to come out nice?! ), add that to the mass of 0.5 kg to get total mass of 2kg, divide by the acceleration (4/9) to get the total force 4.5 N, subtract from 0.5g N to get the friction force 0.405 N, multiply by 0.3 to get the friction torque 1.21 Nm

 

[electro82]

very helpfull tiny-tim

it is a feeling of relief when it all sinks in and it begins to make sense.

again thankyou for your patience you are a saint.

 

[Rafeng404]

Hi all, I too have the same problem and I am suspicious that getting the angular acceleration was that easy... There are pages and pages of examples in my books on how to do it that are much harder. I read somewhere that because it is a flywheel you cannot use alpha = a/r... Is that true?.. Please help haha

 

[tiny-tim]

Hi Rafeng404!

α = a/r isn't physics, it's geometry …

if there's no slipping, then the speed (or acceleration) of the edge of a wheel relative to its centre must be equal to the radius times the angular speed (or acceleration).

If you want further help on this (or anything else), please start a new thread.

(I've noticed that you've posted in several "old" threads … I know some other forums prefer that, but it really doesn't work well in this forum, and particularly not in the homework sections … you'll get more help, and sooner, by starting a new thread)

 

[Mathsishard123]

I am just wondering how you get Ia as 0.199751046 Nm I'm getting 0.01998?