Question

front view side view FT R. ول Pulling the yo-yo [9pt] A non-intuitive motion that combines translation and rotation is pulling a wheel with a string wrapped around it, as shown in the figure above. Depending on what angle the string is pulled, you can obtain three kinds of motion: the yo-yo accelerating in the direction of the applied tension force and winding up, the yo-yo accelerating in the opposite direction of tension and unwinding, and the yo-yo sliding at constant speed and no change in string length. For this problem, assume our yo-yo consists of two identical solid disks each of mass M=0.50 kg and radius R = 30 cm with an effectively massless axle of radius "2R = 15 cm connecting them. A light string is wrapped around the axle, which is also effectively massless compared to the yo-yo disks. The yo-yo sits on a flat table with maximum static and kinetic coefficients of friction us = ulk = 0.80 acting between the wheel and table surfaces. (a) [2 pt] Compute the moment of inertia of the yo-yo about its center of mass. (b) [4 pt] Assume the yo-yo is pulled directly to the left and rolls without slipping. Using Newton's 2nd Laws for translation and rotation, determine the magnitude of tension force and the translational and angular acceleration vectors (magnitude and direction) assuming that the maximum static friction force acts on the yo-yo. (c) [3 pt] Now consider that the string is pulled upward at an angle o relative to horizontal. At what angle would the system be in equilibrium, so that there is neither translational nor angular acceleration?

Answer 1

Na Neemal on each disk ft acm. feciation on each disk FT АЧ Pem R M 0.50kg R = 30cm or 0-3m faf Ms = MK = 0.80 LN Moment of Inutia og Yoyo I about centre of mass I = & Idisc, &E Idisca £MR²+ MR²= MR? 0.5 x 0.3²=0.005 kg me b f = (fmax); us on an=gng a N= Mg 0.5*9.81 4.905N fa 0.8 x 4.905= 3.924 N SlExtref= E(Fic) net m.ac.m. -Fit qf= macm. 7.848 = 2x0.5 acim. ET [F FT - 7.848 = acom. -0

Ełner IL 27xR 0.0452 FT XR Tacoma na 2.35uy 0.15 FTE 0.15 acm. epua FT - 7.848 = 2.3544-0.15 FT 0.15 23.544 LFTE TFT 11.772N acum 3 Fy- 7.848 11.772-7-848 (acoma 3.92u laam. -3.924 こ aum= Ra d = 3.924 13.08 0.3 a = 13.08 i Madis? 2

FT no angular acceleratia F XR T 2 = 2 FAR ang 72f translation acceleration 2N 2 fwso = 24 f=US N 4 Fisino +2N = ang equo [Fr= 4f einh Filoso= 2f 5 54 wo = 2f 어 coso= 1921 Til ficos e FA 4f 0 = Cost (0.5) To = 66