Question

Figure 6 solution see book, solved problem 12.3, also HW problem, also solved during lecture and recitation, also see attachement 14. A thin hollow spherical shell of mass M, and radius R , rolls down without slipping (condition for rolling without slipping :v, = WR) from the top of the incline of length L The incline makes an angle with horizontal ground. What will be the velocity of the centre of mass (C.M) of the sphere when it reaches at the bottom ? L=100 cm, Me 5 kg, 8 = 30° , outer radius R=0.1 m. Moment of inertia of the thin hollow spherical shell about the axis through center of mass is : (2/3)MR (20 points) Hint: find the expression of velocity using conservation of mechanical energy: (19 points) find numerical value of the velocity (1 point) solution see book, section 10.9, also HW problem, also solved during lecture and recitation also see attachment mgh = mgLsind = (1/2)m(VCM)2+(1/2)cm (wa) = (1/2)m(VCM)2+(1/2)cm (Vom/R)2 as VCM = WR November 9, 2017 Con TeXt template 100 m Figure 7 (29Lsind)/(1 + (Icm/MR?)) = on solving we get : VCM = 6gLsin/5 (29Lsind)/(1 + (2/3)) =

Answer 1

4 =Rw 0 Vem : Velocity of center of mass. : " L=100cm = lm M: Sta 0 = 30° w is angulor velocity R = 0.1m radiu of me hollow sphen. g= acceleration clue to gravity = 9.8mi? Sphen is released from rest. So initially it has only Potential. Finally when it reache the bottom of me incline all the energy is converted to Rinche Energy. However, this Kinectic energy is composed of two parts: D Translation Kit due to translakrimal malum y me ceniles { man sphere coresponding to Vem. 2) Rotational Re due to rotation of me sphere about center of man. Therefore By Cow & Consewalions of mechanical Enesgy: (P.C ) intrud + (K.Prompt = (Pelemp + (k.El find 3) Mgh = (k. El final a) mg LSinD (Translal.cored K.elberiman + (Rollionel KE.), ( Opoluhe - Sino

witha = Mgl Sino = M Ver t t Icon Cole ® Mgl sino = Mwind Tem ((X=Rw) Fox xolling million without sipping, lug = Vem 2. Mgl Sind- & mutant de Terme Ver - Mgl SinB = 2(M + I em van Now, for a hollow sphere, I em = 2 MR² . Mgl SinB = 1 (M + 2 MB , vem Multiplying throughout by 2 and discling throughors by M 291 Sino = (+ 3) van - 2gl Sing = (3+2) Ver s van a szalageSine) Vem = Vogl Sino/5 we got

Therefore, Vom - 16 x 9.8 x 1 x Simlo's foto nge- 10 9 Vem = V5.88 = 2.42487 mi! Vem ~ 242 ms) Am