- #1
kritzy
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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?
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?
)