Question

The radius of the roll of paper shown in the figure is 7.6 cm and its moment of inertia is I = 2.9 × 10-3 kg·m2. A force of 2.9 N is exerted on the end of the roll for 1.1 s, but the paper does not tear so it begins to unroll. A constant friction torque of 0.11 m·N is exerted on the roll which gradually brings it to a stop. Assume that the paper's thickness is negligible. (a) Calculate the length of paper that unrolls during the time that the force is applied (1.1 s). 1.8 Incorrect: Your answer is incorrect. m (b) Calculate the length of paper that unrolls from the time the force ends to the time when the roll has stopped moving. m

Answer 1

(a)With the force acting, we write St= Ia about the axis from the force diagram for the roll:
Fr * tfr = Ia1;
(2.9 N)(0.076 m) - 0.11 m · N = (2.9*10^-3 kgm2)a1 ,
which gives a1 = 38.069 rad/s2.
We find the angle turned while the force is acting from
q1 =w0t + ½ a1 t1 2
= 0 + ½ (38.069 rad/s2)(1.1 s)^2 = 23.032 rad.
The length of paper that unrolls during this time is
s1 = rq1 = (0.076 m)(23.032 rad) = 1.75 m.

(b)With no force acting, we write St= Ia about the axis from the force diagram for the roll:
-tfr = Ia2;
-0.11 m N = (2.9*10^-3 kg · m2)a2,
which gives a2 = -37.9 rad/s2.
The initial velocity for this motion is the final velocity from part (a):
w1 =w0 + a1t1 = 0 + (38.069 rad/s2)(1.1 s) = 41.8759 rad/s.
We find the angle turned after the force is removed from
w22=w12+ 2a2q2;
0 = (41.8759rad/s)^2 + 2(-37.9 rad/s2)θ2, which gives θ2 = 23.13 rad.
The length of paper that unrolls during this time is
s2 = rq2 = (0.076 m)(23.13 rad) = 1.76 m.