Question

A solid uniform cylinder of mass M and radius R is at rest in the center on a flat slab of mass m and thickness d, which in turn rests on a horizontal, frictionless table, as shown in the figure at right below. A horizontal force of magnitude F is applied to the slab, acting through the center of the slab, causing it to accelerate to the right. The cylinder rolls without slipping on the slab. (The direction of F is perpendicular to the axis of the cylinder.)

a) Find the magnitude of the friction force Fc between the cylinder and the slab, in terms of M, m, R, and F. (Hint: start by drawing complete free body diagrams of each object. Don’t miss any forces!)

b) Consider the normal force between the table and the slab, which should be on your free body diagram of part (a). Does this normal force act through the center of the slab? Why or why not? And if it does not, calculate the distance ∆ by which it is offset from the center of the slab, and tell in which direction it is offset. (You may answer in terms of Fc, your answer to part (a); no need to plug that expression in.)

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Answer 1

1. mars of cylendo sa Img Normal reaction Acceleration force from table will of block=a, not act at the centre since it can dilma acceleration be part shown by torque balance on block. of centre of Ni - M9 angular acceleration of INI AN (we of cylinder I cylinder a • mutual friction fr. Equation for block > f-for=ma, ol•N, & mg will pass through for cylinder fra ma - . centre of massim of for rotation = fox R = I xa = MR²xx - torque balance werit. centre of mass = Id) - Equation for no slipping of point 'P' (Rolling) * fcx d & Nxtox) = ator = a, (from block) - sz. u acceleration for me. (using er 0,@ 20 ) fc. xd lo hacent per forma ar 3 to a for t sign shows that actually it would E I (ve afo[3, 11= - act on the left side of centre of mass for = F M ANS @ for = fc FEM * N = mg + Mg load = fcxd (Net tor F x = N سام نو I ANS 6 fc 3m + M (pa) = 2(M + m) g ans (6)