Rotational inertia & tangential force

[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.