a. C=pi*D = 3.14 * (2*0.79) = 4.96 cm.
V=5.6cm/s * 6.28rad/4.96cm = 7.1rad/s.
b. C = 3.14 * (2*2.32) = 14.57 cm.
V = 5.6cm * 6.28rad/14.57cm = 2.4rad/s.
In a tape recorder, the magnetic tape moves at a constant linear speed of 5.6 cm/s. To maintain this constant linear speed, the angular speed of the driving spool (the take-up spool) has to change accordingly.
(a) What is the angular speed of the take-up spool when it is empty, with radius r1 = 0.79 cm?
(b) What is the angular speed when the spool is full, with radius r2 = 2.32 cm?
(c) If the total length of the tape is 100.8 m, what is the average angular acceleration of the take-up spool while the tape is being played?
3 answers
C. 0.00294 rad/s^2
C. 100.8 1m*100cm/m /5.6 cm/s = 1800s
(7.1-2.4)rad/s/1800s =.00261 rad/s^2
(7.1-2.4)rad/s/1800s =.00261 rad/s^2