Question

1. (20 points) Two forces F and F, are exerted on a disk with radius R = .5m and a mass M 5kg uniformly distributed F = 30N is exerted upwards and applied a distance R to the right of the disk's center. F = 60N is cxerted downwards and applied a distance R/2 to the left of the disk's center, R/2 E a) What is the aceleration of the center of mass and the angular acceleration around the center of mass if the ground is frictionless? b) What is the acceleration of the center of mass and the angular acceleration around the center of mass if there is friction? You may asume rolling without slipping. c) For this part ignore your answers to parts a) and b) and suppose the angular acceleration around the center is 10rad/s? out of the page. If the disk rolls without slipping and starts from rest how many revolutions will the disk undergo when the speed of the center of mass is 20m/s?

Answer 1

I have uploaded the image showing the answer. I have given b part first then a then c
7 = 12 from figure Tsx 7 = (3/8x F2 + Rx F, - + here în 41 3 الد) .R12 -- Fe CON ķ(5 (55) ginens Fl R = 0.5m 7, Ê (600) Nm = 60 x0.5) E No 15k Nm

T 11 G R o Î (RF) F = 3 ON R = 0.5m Ez Ê (20 x 0.5) Nm Isi Nho 7 = 7, +7 20² ) Nm 7=IZ M = 5kg I = MR? 5*(0.5) ² (kgm?) 2 5x64) kg me & kam?

P= I 2 2-I 30 k I 2 - 240 E - HER 2 - PR In case slipping rolling with out of alcm = ar the direction is the namissa (here) aim - (48x0.5) mist 24 m/s2 Q a) aim and are zero in case fiction Since to move friction is less lase needed. a given x= 10 radle V = 0 V = 20 m/s2

s = distance travelled Vg² - V, 2+Qas Vf, Vo, a are final velocity, initial velocity and acceleration of center y mass then (20) 2 2 Oms a=ar a = 10x0.5 m/s2 = 5m/s2 2x 5 mls?xms 400 (2/2) = =com/52)xs (m) Scm) = 400 co = 4om the wheel will travel a distance 258 one full revolution - 20 R=2 TTX 0.5 am)

total no.y revolutions (40) mo's