Rotational inertia & tangential force

[hage567]

I1 is NOT MR^2. What's the moment of inertia of a solid cylindrical disc? Not that it matters, because you don't need the formula to solve this. In the end you are just going to be solving for I1. So, you're biggest problem right now is figuring out what I2 really means. That's where you have to consider the disc and the putty together.

 

[imy786]

moment of intertia of cylindrical disc= 0.5mR^2
intertia of putty= MR^2

but as u say i don't need this
-----------------------------------------------------

I2= intertia of disc and putty
I1= inertial of disc

intertia of putty= MR^2= 0.007 *R^2
radius=?

not sure if i should choose the radius of 5cm
or the distance of 3cm ?

 

[Doc Al]

intertia of putty= MR^2= 0.007 *R^2
radius=?

not sure if i should choose the radius of 5cm
or the distance of 3cm ?

In the formula MR^2, R represents the distance to the axis of rotation.

 

[imy786]

I2= intertia of disc and putty
I1= inertial of disc

intertia of putty= MR^2= 0.007 *0.03^2= 0.0000063

moment of intertia of cylindrical disc= 0.5mR^2= 0.5*M*0.05^2

I1W1= 12W2

how do i go from here...as i do not know the mass of the disc

 

[Doc Al]

Remember you are trying to solve for the rotational inertia of the disk. (If you knew the mass of the disk, you'd be done in one step!)

If I1 is the rotational inertia of the disk, then I2 = I1 + (rot. inert. of putty).

 

[imy786]

I1W1= 12W2

0.5*M*0.05^2 * 2.666*pi= 0.0000063*2.16667*pi

M= 0.00001024 * 0.05^2
M= 3*10^-8 kg

inertia of the disc= 0.5mR^2
r= 0.05
M= 3*10^-8 kg

interita of disc= o

?

 

[hage567]

I still don't understand why you can't just find I(disc). Why are you still trying to solve for M(disc)? You can't do it!

Anyway, don't forget about moment of inertia of the disc on the right hand side of your equation. You must still take that into account after the putty falls. I mean, there's still a spinning disc involved, right?

So I2 = I(disc) + I(putty) like Doc Al said in post #35.

I think you need to read you textbook a bit more thoroughly, you are missing some basic understanding of this stuff.

 

[Doc Al]

I1W1= 12W2

0.5*M*0.05^2 * 2.666*pi= 0.0000063*2.16667*pi

(1) You do not need the mass of the disk;
(2) You do not need the formula for finding the rotational inertia of a disk;
(3) The final rotational inertia of the system is the sum of the rotational inertia of disk plus rotational inertia of putty; in other words: I2 is not just the rotational inertia of the putty;
(4) The only rotational inertia you need to calculate using a formula is the rotational inertia of the putty.

Do this: Using symbols, rewrite the equation expressing conservation of angular momentum. Use I_d for the disk; I_p for the putty.

 

[imy786]

inertia
-------

owwwww..i think i kinda get it now...

I1W1= I2W2

I2= intertia of disc and putty
I1= inertial of disc

I1* 2.666*pi= 0.0000063*2.16667*pi

I1= 0.00000512

is this correct?

 

[imy786]

i didnt understand y u guys wanted to MERGE THE 2 PARTS OF THE QUESITON...
now its kinda confusing when i want to relate it to the other part anywaz
----------------------------------------------------------------------

 

[imy786]

A constant tangential force is now applied to the rim of the disc
which brings the disc to rest in 8.0 s. Calculate the magnitude of
this force.
--------------------------------------------------------------------

T= I *alpha

I= 5.12*10^-6* angular accelreation

angular accelration= change in angular vecloity / time

angular vecloty= 65/rev / min
= 2.166pi rad/ sec

angular accelation= 0.850

T= I *alpha

I= 5.12*10^-6* 0.850
= 4.36*10-7

 

[hage567]

inertia
-------

owwwww..i think i kinda get it now...

I1W1= I2W2

I2= intertia of disc and putty
I1= inertial of disc

I1* 2.666*pi= 0.0000063*2.16667*pi

I1= 0.00000512

is this correct?

No, I2 is still wrong. I2=I(disk)+I(putty). You never changed it from what it was before.
Do it algebraically, without numbers. There is a I(disc) term that you must move form the right side to the left when you do it properly.

 

[hage567]

i didnt understand y u guys wanted to MERGE THE 2 PARTS OF THE QUESITON...
now its kinda confusing when i want to relate it to the other part anywaz
----------------------------------------------------------------------

You need the first part to get the second part. It helps for references purposes to have all the information in one place, not scattered all over the forum.

 

[hage567]

A constant tangential force is now applied to the rim of the disc
which brings the disc to rest in 8.0 s. Calculate the magnitude of
this force.
--------------------------------------------------------------------

T= I *alpha

I= 5.12*10^-6* angular accelreation

angular accelration= change in angular vecloity / time

angular vecloty= 65/rev / min
= 2.166pi rad/ sec

angular accelation= 0.850

T= I *alpha

I= 5.12*10^-6* 0.850
= 4.36*10-7

Why are you solving for I, when you put I in as a variable in the line above it? Is that just a typing error? You are supposed to use I2 for this part, which has been defined for you many times above.

You are trying to find the force applied, which as was discussed earlier, is not the same as torque.

 

[imy786]

I1W1= I2W2

I2= inertia of disc and Inertia of putty
I1= inertia of disc

I1w1= I2w2

I(disc)w1 = [I (disc) + inertia (putty)] w2

w1/w2 = [I (disc) + inertia (putty)] / I disc

w1/w2 = 1 + [I (putty) / I disc]

w1= 2.6666*pi
w2= 2.1666*pi
intertia of putty= MR^2= 0.007 *0.03^2= 0.0000063

2.6666/2.1666 = 1+ [0.0000063 / I disc]

I disc= 2.73*10^-5

 

[Doc Al]

Looks good. (But don't forget units.)

You could save yourself a bit of work by immediately isolating I(disc) in your equation; starting form this:

I(disc)w1 = [I (disc) + inertia (putty)] w2

You can solve for I(disc) like so:
I(disc)w1 = I(disc)w2 + I(putty)w2
I(disc)(w1 - w2) = I(putty)w2
I(disc) = I(putty)w2/(w1 - w2)

(Which is, of course, equivalent to what you did; but this seems a tad easier to me.)

 

[imy786]

wowwwww i can't belvue after all that i did it right?

this is on of the most confusing quesitonsi kinda ever done...

now...to work out the force...
------------------------------------------------------------
well as everyone keeps telling me force isn't TORqu

A constant tangential force is now applied to the rim of the disc
which brings the disc to rest in 8.0 s. Calculate the magnitude of
this force.
----------------------------------------------------------
force= mass* angular accelreation

Torque= I*angular acceleration

from 65 rev/min to ZERO in 8 seconds!

65 rev/ min= 130pi rad/ min
2.166667 rad / sec

I disc= 2.73*10^-5

Torque= I*angular acceleration

2.73*10^-5*2.166667= 5.91498 N

is this correct...?

 

[Doc Al]

now...to work out the force...
------------------------------------------------------------
well as everyone keeps telling me force isn't TORqu

Force isn't torque. Force and torque are two different--but related--quantities. Torque is a force times perpendicular distance to the axis. (Read this: http://hyperphysics.phy-astr.gsu.edu/hbase/torq.html#torq")

The first step is to find the torque, then use the torque to find the force.

Torque= I*angular acceleration

Good. This is Newton's 2nd law for rotation.

from 65 rev/min to ZERO in 8 seconds!

65 rev/ min= 130pi rad/ min
2.166667 rad / sec

What did you calculate here? You already now w2 from earlier--now you need the angular acceleration.

Redo this step and then find the torque.

(But don't stop there--you need the force, not just the torque.)

 

[imy786]

A constant tangential force is now applied to the rim of the disc
which brings the disc to rest in 8.0 s. Calculate the magnitude of
this force.
--------------------------------------------------------------

as the disc comes to rest = meaning 0 rad/sec ,dont i need to calculate angular accelration

w2= 2.1666*pi
angular accelration= angular veloctiy/ time
= 2.166667pi/ 8=0.271 rad/sec^2

torque= I*a
= 2.73*10^-5*0.271
=7.39*10^-6

toqrue= force*distance
7.39*10^-6= f*d

d=? not sure what the distance would be, the force is applied on the rim of the disc.

 

[hage567]

w2= 2.1666*pi
angular accelration= angular veloctiy/ time
= 2.166667pi/ 8=0.271 rad/sec^2

This answer isn't right, you dropped pi in your calculation.

I disc= 2.73*10^-5

I think you are supposed to be using the moment of inertia of the disc plus the putty, since the putty is still on the disc when the force is applied.

d=? not sure what the distance would be, the force is applied on the rim of the disc.

Look at the question. It tells you this.

 

[imy786]

i have been given the diamter = 10cm and the distance 3cm...

i would asssum distance 3cm. (or would i have to do 10-3).

w2= 2.1666*pi
angular accelration= angular veloctiy/ time
= 2.166667pi/ 8=0.85 rad/sec^2

torque= I*a
= 2.73*10^-5* 0.85 = 2.32*10 ^ -5 Nm

toqrue= force*distance
2.32*10 ^ -5= f*0.03

f=7.7*10^-4 N

 

[hage567]

Well if the force is being applied to the edge of the disc, and the disc is 10 cm in diameter, why would you put the distance as 3 cm? 3 cm is the distance to the putty, right?

You are not using the right value for I. 2.73x10^-5 is the value for the disc alone. If the putty is on the disc, you must add the two individual I values to get the total moment of inertia (we've been through this idea already). So the total would be 2.73x10^-5 + I(putty). I(putty) you have found already. Do you understand what I mean?

 

[imy786]

i understand...just that i was looking at other parts of this problem that i missed the I (putty ) out.

d= 10cm = 0.1 m

torque= I*a

I= I disc + I putty
= 2.73x10^-5 + 0.0000063
= 3.36*10 ^-05

torque= I*a
= 3.36*10 ^-05*0.85= 2.856*10 ^-5

T= Fd
2.856*10 ^-5 / d = F
2.856*10 ^-5/ 0.1 = F
=2.856*10^-4 N

 

[hage567]

One more thing: the distance is not 0.10 m. Since the torque is about the centre of the disc, you want to use the radius of the disc. That is the distance from the axis of rotation to where the force is being applied.

Other than that, I think everything looks OK.

 

[imy786]

distance would be 5cm...then right?

0.5 m

 

[hage567]

5 cm is not 0.5 m. Try again. Remember, there is 100 cm in a meter.

 

[imy786]

5cm= 0.05 m

 

[imy786]

(part c)

Calculate the rotational energy of the system before and after the putty is
added to the disc. Comment briefly on your answer.
------------------------------------------------------------------------

formulas that can be used
Erot= 0.5*I*w^2
Erot= 0.5*m*r^2*w^2


attempted solution:

Erot b4 putty is added= Erot only of disc=0.5*I*w^2


I disc= 0.0000063
w= 2.6667*pi

Erot= 2.2*10^-4 J is the rotational energy of the disc before the disc is added.


-----------------------------

Erot after the disc is added:

Erot= 0.5*I*w^2

I disc= 2.73*10^-5
I of putty =0.0000063

I of putty + I of disc
0.0000063 + 2.73*10^-5= 3.36*10 ^-05

w= 2.1666*pi

Erot=24.8*10^-5 J after the disc is added

 

[hage567]

formulas that can be used
Erot= 0.5*I*w^2
Erot= 0.5*m*r^2*w^2

Be careful, this isn't always the case. It depends on the geometry of the object.

Attempted solution:

Erot b4 putty is added= Erot only of disc=0.5*I*w^2

I disc= 0.0000063 This is not I for the disc! You found that to be 2.73x10^-5, remember?
w= 2.6667*pi OK

Erot= 2.2*10^-4 J is the rotational energy of the disc before the disc is added. ? This statement makes no sense!
This answer is not right.

You are still getting things seriously mixed up. I suggest you make a clear list of all the variables so you know which is which.

Erot after the disc is added: Again, this should say putty.

Erot= 0.5*I*w^2

I disc= 2.73*10^-5 This one is right.
I of putty =0.0000063 So it this one.

I of putty + I of disc
0.0000063 + 2.73*10^-5= 3.36*10 ^-05 Good here.

w= 2.1666*pi OK

Erot=24.8*10^-5 J after the disc is added NO, the putty!
I don't see how you got this number.