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Discuss the effect of rolling on inclined plane and derive the expression for the acceleration. |
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Answer» Solution :Acceleration of the rolling object : (i) From the figure, it is seen that the component of gravitatinal force (mg cos`theta`) perpendicular to inclined plane is balanced by the normal force N. Therefore, the component of GRAVITATIONAL force (mg sin`theta`) parallel to inclined plane and the frictional force f, combinely causes the motion. (iii) For translational motion, `mg sin theta - f = ma""...(1)` (IV) For rotational motion, let us take a torque with respect to the center of the object. mg sin`theta` cannot make any torque as it PASSES through the center of the object, but the frictional force f can set a torque as, `tau = Rf` (v) But we know, `tau = I alpha`, thus, `Rf = I alpha` (vi) Substituting, the ANGULAR acceleration `alpha = (a)/(R)` and the moment of inertia `I = mK^(2)`, we get, `Rf = mK^(2) (a)/(R)` `f = ma ((K^(2))/(R^(2)))""...(2)` (vii) Substituting equation (2) in (1), we have, `mg sin theta - ma ((K^(2))/(R^(2)))=ma` `mg sin theta = ma + ma ((K^(2))/(R^(2)))` `a (1+(K^(2))/(R^(2))) = g sin theta` (vii) After REWRITING it for acceleration, we get `a = (g sin theta)/((1 + (K^(2))/(R^(2))))` |
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