Question

A hollow sphere of radius 0.240 m, with rotational inertia I = 0.0470 kg·m² about a line through its center of mass, rolls without slipping up a surface inclined at 13.3° to the horizontal. At a certain initial position, the sphere's total kinetic energy is 12.0 J. (a) How much of this initial kinetic energy is rotational? (b) What is the speed of the center of mass of the sphere at the initial position? When the sphere has moved 0.690 m up the incline from its initial position, what are (c) its total kinetic energy and (d) the speed of its center of mass?

Answer 1

Answer Griven: R. 0.24 m 2 0.047 kg m 13-3 12 J. 13.3" 0.69m d a) (Kinetic energy) f for (k-Evo (Kokavan Tw" + Mv 2 2 2 1 (K. E) hot 2 2 3 * ? MRWE IM(RW) VaRW) I a 2 MR² and 3 1 (K. E) MR w² + 1 MR² w² tou 3 2 1 MR²+ I MR 3 2 2 ule know that: I 3I 2 MR 3 - Mass, M: 2R2 3 x 0.041 1.224 kg 2x0.2012 Substituting in equation ® 12 x 1-224*(0-24) + 1 x 1.220 * 0.242 2 3 w ² = 204.25 204-25 = 14.29 yall's

(K. ) vor 1 I w² 2 2 1 x 0.047 x 2 4.8 I b) Speed: V. Ro 0.24 X 14.29 3.43 m/s from conserva Hon of mechanical energy (K. E) fot K E+ P.E = KE + Mg b. hd sino (133) 0.69 x sio (13.3) 0.159 m kot (ko 8 tot Mgh 12 ( 1-224 x 9.8 x 0.159) 10.09 J from O > w I MR + I MR? 3 2 K.E 5 MR² af Rw K.F 5 M. 6

? : KE 5 M. K 10.09. 프M 5 x 1.214. 3-145 mls 다.