Calculate torque required to accelerate question

[ingram010]

Hi all

getting a bit stuck on this problem:-

a solid cylinder rotating about its polar axis has a mass of 120kg and a diameter of 0.6,. If the bearings provide a frictional torque of Nm, find the torque applied to accelerate the cylinder from 200 to 1400rpm in 40 seconds?

if anyone can provide the formula to solve this I will be very grateful

cheers

John

 

[tiny-tim]

hi ingram010!

first use one of the standard constant acceleration equations to find the angular acceleration

then use torque = Iα ​

 

[ingram010]

Thanks tiny-tim

I will take a look

 

[ingram010]

Hi Tiny-Tim

I have solved the problem but I can’t work out how to factor in the loss of torque due to friction.

Is it just a case of deducting it from the applied torque?

Hope you can help

Cheers

John

 

[tiny-tim]

Hi John!

… I can’t work out how to factor in the loss of torque due to friction.

Is it just a case of deducting it from the applied torque?

That's right!

Torque, like any vector (ok, strictly it's a pseudovector ), is additive …

you found the angular acceleration, that tells you the total torque, so just subtract the friction torque to get the applied torque.

 

[ingram010]

Great

so say the applied torque is 33.93Nm I just simply subtract the frictional torque which is 1Nm giving 32.93Nm?

It seems too simple to be true, but Ill go with it : )

Thanks for you help

 

[Quinzio]

Excuse me, why do you want to SUBTRACT ?
You need a force (torque) F to accelerate something. If you have a friction you'd need an ADDITIONAL force.
Do you agree ?

Are you sure about 33.93 Nm ?

 

[ingram010]

Oh yes, whoops.

I have used the following equations to solve the problem:-

200rpm x 2 pi /60 = 20.94 rads/s = initial velocity
1400rpm x 2 pi /60 = 146.61 rads/s = final velocity

then

final velocity - initial velocity/40 seconds = 3.142 rads/s^2 = angular acceleration

then

mass x acceleration x radius^2 = Torque

so 120kg x 3.142 x 0.3^2 = 33.93Nm + frictional torque of 1Nm = 34.93Nm

Am I doing something wrong? I have to get this nailed as I have an exam on monday.

 

[Quinzio]

mass x acceleration x radius^2 = Torque

Did you mean
$$\tau = I \alpha$$

torque = acc x inertia ?

So revise the cylinder inertia

 

[tiny-tim]

… mass x acceleration x radius^2 = Torque

so 120kg x 3.142 x 0.3^2 = 33.93Nm + frictional torque of 1Nm = 34.93Nm

it would help you if you wrote things out properly

acceleration = 3.142

so total torque = 33.93

but total torque = applied torque minus friction torque

so applied torque (as asked for in the question) = …

(but your moment of inertia is wrong)

 

[ingram010]

According to the book I have, inertia = mass x radius^2

so I = mass x raduis^2 = 10.8kgm^2

T= I x acc = 10.8 x 3.142 = 33.93

and
mass x acc x radius^2 = 120kg x 3.142 x .3^2 = 33.93

it gives the same answer, I am not sure what you are telling me

John

 

[tiny-tim]

According to the book I have, inertia = mass x radius^2

(it isn't inertia it's https://www.physicsforums.com/library.php?do=view_item&itemid=31")

see https://www.physicsforums.com/showthread.php?t=503084&goto=newpost" for formulas which you need to learn

 

[ingram010]

this is the extract from the book I am using:-

Torque, moment if inertia and angular motion

Torque

From Newton's third law, we know that to accelerate a mass we require a force such that:

F= ma

Now in dealing with angular motion, we know that this force would be applied at a radius r, from the centre of rotation and would thus create a turning moment or more correcxtly a torque T, thus

T = Fr or T = mar

Since the linear acceleration , a = rα(angular acceleration), then

T = m(rα)r or T = mαr^2.

Moment of inertia and angular motion

Now from our equetion for torque the quantity mr^2 has a special significance. It is known as the moment of inertia of the body about its axis of rotation. It is given the symbol I, thus: I = mr^2 and the units of I are (kgm^2), this is because the inertia of a body, from Newton's first law, is proportional to its mass multiplied by the distance squared.

Quite rightly I should have called it moment of inertia, but I still can't see what I am diong wrong.

the revised statement should read

According to the book I have, moment of inertia = mass x radius^2

 

[ingram010]

Ahh!

Just found this, I really should spend more time reading.

Moment of inertia for a solid disk = 1/2 mr^2

 

[ingram010]

Hi Tiny-Tim and Quinzio

I am sure that the answer to my question is 17.97Nm

Thank you both very much for your help

Kindest Regards

John
p.s If I am wrong please tell me : )

 

[Quinzio]

Ok..