A twine-spooling operation at a gift-wrap factory uses a motorized wheel of 7 cm

Question

A twine-spooling operation at a gift-wrap factory uses a motorized wheel of 7 cm radius and 500 gram mass to wind twine onto spools. One such wheel is rotating at 1400 rpm when the motor’s power supply fails, and friction starts to slow the wheel. By the time the workers realize what has happened, 119.3 m of string have been improperly wound and the wheel is turning at a rate of 1050 rpm.

a) What is the magnitude of the average angular acceleration provided by friction?

b) How long did it take for the wheel to slow to 1200 rpm?

c) By the time the wheel stops, how much string will need to be unwound from the spool?

Explanation / Answer

let
wi = 1400 rpm

= 1400*2*pi/60

= 146.6 rad/s

angular displacement, theta = s/r

= 119.3/0.07

= 1704 rad

wf = 1050 rpm

= 1050*2*pi/60

= 110 rad/s

a) use, wf^2 - wi^2 = 2*alfa*theta

==> alfa = (wf^2 - wi^2)/(2*theta)

= (110^2 - 146.6^2)/(2*1704)

= -2.76 rad/s^2

b) wf = 1200 rpm

= 1200*2*pi/60

= 125.7 rad/s

time taken to slow down,

t = (wf - wi)/alfa

= (125.7 - 146.6)/(-2.76)

= 7.57 s

c) maximum angular displacement, theta = (wf^2 - wi^2)/(2*alfa)

= (0^2 - 146.6^2)/(2*(-2.76))

= 3893 rad

use, s = r*theta

= 0.07*3893

= 272.5 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.