Question

4:36 # $24 OKO 34% A solid cylinder with mass m and radius r rolls without slipping down an incline that makes a 3B" with the horizontal What is the moment of inertia of the solid cylinder about the center of mass? Incorrect I = 2/5.mp Incorrect I = 1/12 + m2 Correct: I = mr/2 Incorrect T = mr2 You are correct. Your receipt no. is 158-399 Previous Tries What form does Newton's 2nd Law take in this system (F is the force due to friction)? Take the positive direction to be down the ramp Incorrect F = mg - F, = ma Incorrect F = mg sino - F, = 0 Incorrect SF = mg. cos - F, = ma Correct: F = mg.sin - F, = ma You are correct. Your receipt no is 158-1967 Previous Tres What is the sum of the torques about the center of mass of the cylinder? In case the font doesn't render clearly, 'a' is 'alpha', the angular acceleration of the cylinder Incorrect T = F, T.sin = lo Incorrect T = F, = 0 Incorrect T = mg. = la Correct: = F = la You are correct Your receipt no. is 158-982 Previous Tres What is the acceleration of the cylinder? Submit Answer Tries 0/5

Answer 1

solid sptere cylinder Moment of Imertia of IrmR21-0 mgsino * In parallel to planet | ng sino - fs =ma- Torque = fer = .— For pure rolling az a fs= Ia - substituting ♡ unto 6 mysino fos I a = ma (m + 2 ) a= mg sino R2 sino - 9.8 x sin 380 mk- a = x 9.8 x sin 38° = 4:02 m/s2|