4. A roll of toilet paper (a partially hollow cylinder with and I 90x 10 m is R2

Question

4. A roll of toilet paper (a partially hollow cylinder with and I 90x 10 m is R2 7.5 cm, M 300. g, mounted on a massless axle along its central axis. The roll is initially at rest. Then, at t J0, a rascally child grabs the end of the roll and starts running, pulling the paper off the roll at a constant linear acceleration atan 0.35 m/s2 Assume that the roll has uniform density. Throughout this question, assume that both M and R2 remain constant, even though paper is unspooling from the roll Be sure to include correct units! a. (2 pts.) What is the spool's inner radius, Ra? b. (3 pts.) What is the magnitude of the torque acting on the spool? c. (2 pts.) At what time does the spool reach a rotation rate of 400. rpm? d. (2 pts.) The spool holds only 40. m of paper. If the child maintains the same constant acceleration. at what time will the spool run out of paper? Even though paper is unspooling from the roll, assume that R2 and Mremain constant. Bonus (+1 pt): At the moment that the spool runs out of paper, how manyrevolutions has it executed? Even though paper is unspooling from the roll, assume that R2 and Mremain constant.

Explanation / Answer

a)I = 1/2*(R12+R22)M

putting M and R1 we get

R2 = 0.01936 m or 1.936 cm

b) Torque = m*a*R1 = 0.3*0.35*7.5/100 = 7.875*10-3 N-m

c)angular acceleration = 0.35/0.075 = 4.666 rotations/sec^2

rotation rate = 400rpm or 6.666 rpsec

use formula

v = u +a*t

6.666 = 4.666*t

so t = 1.428 secs

d) circumference of spool = 2*(22/7)*7.5/100 = 0.4714 m

for covering 40 m it will have to round 40/0.4714 rotations or 84.85 rotations

use formula

s = u*t +1/2*a*t2

84.85 = 1/2 *4.666*t2

thus t = 6.03 secs

answer of BONUS question is in the above solution only

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