Question

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Problem 2 (10pt). A uniform disk with mass m = 8.55 kg and radius R= 1.35 m lies in the xy plane and centered at the origin. Three forces act on the disk in the +y-direction (see figure below): (1) a force Fi = 335 N at the edge of the disk on the +x-axis, (2) a force F2 = 335 N at the edge of the disk on the -y-axis, and (3) a force F3 = 335 N at the edge of the disk at an angle 0 = 30° above the - x-axis. F2 Figure 2: Problem 2.

(a) What is the magnitude of the torque on the disk (about the z axis) due to F1? (1pt)

(b) What is the magnitude of the torque on the disk due to F2? (1pt)

(c) What is the magnitude of the torque on the disk due to F3? (1pt)

(d) What is the magnitude and direction (clockwise or counterclockwise) of the angular ac- celeration about the z-axis of the disk? (2pt)

(e) If the disk starts from rest, what is the angular speed and the rotational kinetic energy of the disk after the forces have been applied for t = 0.6 s? (2pt)

(f) What is the linear acceleration and velocity of a point on the rim of the disk at 0.6 s? (2pt)

(g) How many revolutions does the disk make during first 0.6 s? (1pt)

Answer 1

Giren 'F=3350, F2 = 335 N naty © due to F3? Crs - x Con30% (5 = 1.35*335 C430 = 391.66 N ④ - Into the page) (-2) (d Tuet = 452-39466 ④ = 60.34 7 Hence Moection of Anguly acc is 7=335N, F 335 N County Clock-wire (E=12). and 5=835N, O=30° - Z -> Counter cloch were 1 I = I MR² m=855rg, R=1.350 =48.5571.35 = 7.79 kg, m @ Toome che to fi 12= 7 = 7.74 rad/ do kaugules acct. = 1:35x335 x singo Hence anguly speed a = 452125 N-m 45025 Nm wa u W = Wotant = xt = 7.74x0-6 rw=4.646 rad/ As 6 Tome due to fa :- - F - 1.354335 sin 180 KE = 1 IW² = 1x7.79 x 4.648 If = 0 N-m Ay IKE = 84.10 J. Ars & Lineer acceleration (a)=2R = 7.74 +135=10.449 m/g² As Liner relocity na wR=4.646x1135= 6.2721 m/s Ay o no. of revulation 0 = 1 1 2 2 + ² = 1 x 7.744.6² = 1.393 rade Ay INLOVE