Rotational inertia & tangential force

[imy786]

Homework Statement

A horizontal disc of diameter 10 cm is spinning freely about a vertical axis
through its centre at an angular speed of 80 revolutions per minute. A piece
of putty of mass 7.0 g drops on to and sticks to the disc a distance of 3.0 cm
from the centre. The angular speed reduces to 65 revolutions per minute.

Calculate the moment of inertia of the disc.

Homework Equations

I= MR2

The Attempt at a Solution

R= 0.03
M= 0.007

but if i use the formula I= MR2,,,, what about the informaton they have given of rev/ min this is the angular speed...

not sure the method to do this?

 

[Doc Al]

conservation

When the putty is dropped onto the rotating disc, some things change and some things remain the same. What stays the same?

 

[imy786]

mass remains same, diamiter remains the same
angular velocity decreases

 

[Doc Al]

what's conserved?

mass remains same, diamiter remains the same
angular velocity decreases

All true, but what important dynamical quantity is conserved?

 

[SGT]

mass remains same, diamiter remains the same
angular velocity decreases

Mass of the disk remains the same, but the mass and moment of inertia of disk + putty is not the same.

 

[imy786]

so calculating this...

is my formula corect?

 

[hage567]

What formula are you referring to?? I=MR^2 is correct for the putty if that's what you mean. But you haven't yet taken into account the moment of inertia of the disk.

 

[imy786]

mass remain the same-

how do i calcualte the moment of inertia...with the informaiton given>>

 

[Doc Al]

mass remain the same-

More importantly, angular momentum remains the same. Compare the angular momentum of the system before and after the putty is dropped.

how do i calcualte the moment of inertia...with the informaiton given>>

Since the moment of inertia of the disc is what you are trying to find, just call it I_d. But realize that the moment of inertia of "disc + putty" is the sum of the individual moments of inertia. You can calculate the moment of inertia of the putty using I_p = MR^2.

 

[imy786]

tangential force

Homework Statement

A horizontal disc of diameter 10 cm is spinning freely about a vertical axis
through its centre at an angular speed of 80 revolutions per minute. A piece
of putty of mass 7.0 g drops on to and sticks to the disc a distance of 3.0 cm
from the centre. The angular speed reduces to 65 revolutions per minute.

Homework Equations

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.

The Attempt at a Solution

need help..dont know how to start

 

[cristo]

What do you know about circular motion? You are given the angular speed of the disc, and the time in which it is brought to rest. Can you see how this will help you find the force?

Please show some work, even if it's just relevant equations/ knowledge from your text or course notes. You must know how this forum works by now!

 

[imy786]

didnt have a clue wear to start,,thats for the tip...

f=m*r*w^2

= 7000*0.05*w^2

w= 80-65 / 2= 7.5

f= 3.5*7.5^2
= 196.875N

 

[imy786]

can anyone tell me if this correct?

 

[hage567]

No, that doesn't look right.

How did you get 7000 for the mass? You shouldn't even be using mass here, you should be using moment of inertia, I. Why are you solving for w? You know that already. You're dealing with angular momentum and torque, right? How do you think you can relate things like force, torque, and angular acceleration?

Also, be careful of your units.

 

[imy786]

torque= force *lever arm

angular acceleration= a/r

a=acceleration
r=radius

a= v-u/ t

change in velocity= 65-80= -15 rev/ min

-15 rev/min= -15 rev every 60 sec = -0.25 rev/ sec

-0.25/8= -0.03125 rev/sec^2

angular acceleration= -0.03125 rev/sec^2

tangential force= mass*angular acceleration
= 0.07 * -0.03125
0.0021875 N

this answer doesn't look right.
Need help please.

 

[hage567]

torque= force *lever arm

angular acceleration= a/r

a=acceleration
r=radius

a= v-u/ t

change in velocity= 65-80= -15 rev/ min

-15 rev/min= -15 rev every 60 sec = -0.25 rev/ sec

-0.25/8= -0.03125 rev/sec^2

angular acceleration= -0.03125 rev/sec^2

tangential force= mass*angular acceleration
= 0.07 * -0.03125
0.0021875 N

this answer doesn't look right.
Need help please.

OK first of all, the disc is coming to a STOP when the force is applied, so that means your final angular velocity is ZERO. Take the time to look at what the question is asking you!

It is best to convert angular velocities into rad/sec.

You STILL didn't convert 7g to kg properly, but you need to consider more than just the mass, you are dealing with rotational quantities.

Remember that torque can be expressed as

\tau = I\alpha

where I is the moment of inertia. That is what you need to consider here (I pointed this out in my last post as well). Use this along with the first equation you listed. I know there was a first part to this question that you posted before, so you should already know the moment of inertia you need to use to solve this.

 

[imy786]

7g= 7kg/1000
= 0.007kg

Torque= I*alpha..

what is alpha...

the question also ask to calcualte tan...force...why do we need torque?

 

[Hootenanny]

Torque= I*alpha..

what is alpha...

alpha is the angular acceleration.

the question also ask to calcualte tan...force...why do we need torque?

What is torque...

 

[hage567]

alpha is the angular acceleration. The force is creating a torque on the disc that is bringing it to a stop.

 

[imy786]

is torque a force...

if so,

then

torque= I* alpha

I= MR^2

angular accelereation= change in angular velocity/ time
= (65- 80) / 8= -1.875 rev/sec^2

 

[Hootenanny]

is torque a force...

Torque is not a force. A torque is produced when the applied force does not pass through the centre of mass of the body.

I= MR^2

This is the moment of inertia fo a disc; however, the disc has a point mass located 0.03m from the axis of rotation. You must take this into account and calculate the moment of inertia for the composite body.

 

[hage567]

is torque a force...

if so,

then

torque= I* alpha

I= MR^2

angular accelereation= change in angular velocity/ time
= (65- 80) / 8= -1.875 rev/sec^2

WHY are you STILL using 65 and 80?!

The question says:

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.

That means it goes from 65 rev/min to ZERO in 8 seconds!
You are NOT supposed to be finding the deceleration for when the putty falls on the disc, if that's what you're doing.

Also realize that the units of the angular speed are given in rev/min. The time interval the disc take to stop is 8 seconds. Your units are not consistent and make no sense.

 

[imy786]

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

angular momentum is converved

disc-
M=?
R=0.12m

moment of interia of disc=0.0072M


putty-
M=0.005
R=0.4

moment of inertia of putty= 0.000008


--------------
need help if i can do more to calculate the moment of inertia of disc.

 

[Hootenanny]

Note to mentors: Perhaps we could merge these two threads https://www.physicsforums.com/showthread.php?t=168385

 

[hage567]

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

angular momentum is converved

disc-
M=?
R=0.12m

moment of interia of disc=0.0072M


putty-
M=0.005
R=0.4

moment of inertia of putty= 0.000008


--------------
need help if i can do more to calculate the moment of inertia of disc.

You need to use conservation of angular momentum to find the moment of inertia of the disc. You still haven't done that. You don't have to find the mass of disc explicitly to solve this problem.

So if angular momentum is L = I\omega, for conservation you must have

L_i=L_f

so I_i\omega_i=I_f\omega_f

So initially, I_i only consists of the moment of inertia of the disc, I_d. Once the putty hits the disc, what will be the new total moment of inertia?

Read Doc Al's post carefully (#9).

 

[Doc Al]

Note to mentors: Perhaps we could merge these two threads ...

Threads merged. Note to imy786: Don't start multiple threads on the same problem. Don't start a thread about part 2 of a problem when you haven't finished part 1 (especially when you need to solve part 1 first).

 

[imy786]

can someone check last answer..

 

[hage567]

can someone check last answer..

It's wrong. None of the numbers you used in post #23 for mass or radius are given in the question, so where in the world did you get them from?

 

[imy786]

L= Iw

angular speed=w= 72 rev/ minute...need to change this rad/sec

72*2pi / min
144pi/ min
= 144pi/ 60 secs
=24pi/10 rad/sec

I1W1= I2w2

w1= 2.4 pi rad/sec
w2=2pi rad/sec

I1 = MR^2

 

[hage567]

L= Iw

angular speed=w= 72 rev/ minute...need to change this rad/sec

72*2pi / min
144pi/ min
= 144pi/ 60 secs
=24pi/10 rad/sec

I1W1= I2w2

w1= 2.4 pi rad/sec
w2=2pi rad/sec

I1 = MR^2

Where are these values you keep using coming from? There is no 72 rev/min in this question! LOOK at the question you posted!

 

[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.