Unrolling of Paper (Torque/Rotational Motion)

[kritzy]

1. The radius of the roll of paper is 7.9cm and its moment of inertia is I= 3.1×10−3 kg m^2 . A force of 2.3 N is exerted on the end of the roll for 1.3 s, but the paper does not tear so it begins to unroll. A constant friction torque of 0.13 mN is exerted on the roll which gradually brings it to a stop. (Part A)Assuming that the paper's thickness is negligible, calculate the length of paper that unrolls during the time that the force is applied (1.3s). (Part B)Assuming that the paper's thickness is negligible, calculate the length of paper that unrolls from the time the force ends to the time when the roll has stopped moving.

2.Relevant Equations
$$\tau$$=RF
$$\tau$$=I$$\alpha$$
a=R$$\alpha$$
x=$$\frac{1}{2}$$at$$^{2}$$

3. The Attempt at a Solution [/b]

I don't understand torque and rotational motion very much but here's my attempt.
(Part A) Since they give a force, I found the torque.
$$\tau$$=RF
(.079)(2.3)=.1817
Then I find the sum of torques.
.1817-.13=.0517
$$\Sigma$$$$\tau$$=I$$\alpha$$
$$\alpha$$=$$\frac{\Sigma\tau}{I}$$
$$\alpha$$=.0517/.0031=16.6
a=R$$\alpha$$=(16.6)(.079)=1.3
x=.5at^2=(.5)(1.3)(1.3)^2=1.0985
I don't think that's right. Unfortunately, I don't understand Part B either. Can somebody please lead me in the right direction?

 

[tiny-tim]

Welcome to PF!

(Part A) Since they give a force, I found the torque.
$$\tau$$=RF
(.079)(2.3)=.1817
Then I find the sum of torques.
.1817-.13=.0517
$$\Sigma$$$$\tau$$=I$$\alpha$$
$$\alpha$$=$$\frac{\Sigma\tau}{I}$$
$$\alpha$$=.0517/.0031=16.6
a=R$$\alpha$$=(16.6)(.079)=1.3
x=.5at^2=(.5)(1.3)(1.3)^2=1.0985
I don't think that's right. Unfortunately, I don't understand Part B either. Can somebody please lead me in the right direction?

Hi kritzy ! Welcome to PF!

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

Your Part A looks ok to me … what's worrying you about it?

For Part B, first calculate the velocity (or angular velocity) when the 2.3N stops, then use the new (negative) acceleration in the appropriate constant acceleration equation to find the distance until the velocity is zero

 

[kritzy]

Hi kritzy ! Welcome to PF!

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

Your Part A looks ok to me … what's worrying you about it?

For Part B, first calculate the velocity (or angular velocity) when the 2.3N stops, then use the new (negative) acceleration in the appropriate constant acceleration equation to find the distance until the velocity is zero

I understand now! Thank you for your help. I thought my answer for part A was wrong so I was too scared to submit it. But it worked out okay. By the way, how do I mark the thread as solved? Thanks again.

 

[tiny-tim]

I understand now! Thank you for your help. I thought my answer for part A was wrong so I was too scared to submit it. But it worked out okay. By the way, how do I mark the thread as solved? Thanks again.

Hi kritzy!

The "solved" facility disappeared during the Great Server Update of '08.

I'm glad it all worked out.

See you around. ​