A bobbin has inner radius r and outer radius R is placed on a rough horizontal surface. A light is string wrapped over inner core, connects a block with bobbin as shown in the figure. Now system is released from rest and bobbin moves on the horizontal surface without sliding, and the string does not slide from bobbin, If rR=0.25, find the ratio of the acceleration of the block and bobbin aA.

As the motion of the bobbin is pure rolling, point P should be instantaneous centre of rotation (ICR) of the bobbin. Hence we can consider pure rotation of the bobbin about P. The acceleration of point C; aC=αR.But the bobbin is moving with acceleration A,Hence A=αR            …(i)The acceleration of point Q; aQ=α(R−r).But point O is directly connected with block, it means the magnitude of acceleration of the point Q should be equal to the acceleration of the blockaQ=a=α(R−r)         …(ii)From (i) and (ii), we getaA=R−rR=1−rR=1−0.25=0.75