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| Moment of Inertia of Circular Ring, Disc, Cylinder, Sphere and Thin Straight Rod |
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Consider a solid cylinder of mass M and radius R, rolling down a plane inclined at an angle  with the horizontal. Let us assume that the cylinder rolls without slipping. The condition for rolling without slipping is that at each instant, the point of contact P (rather, the line of contact) is momentarily at rest and the cylinder is rotating about that as axis. The centre of mass of the cylinder moves in a straight line. |
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| The external forces acting on the cylinder are as follows: |
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 The weight Mg of the cylinder, acting vertically downwards through the centre of mass of the cylinder |
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 The normal reaction N of the inclined plane, which is equal to Mg cos q |
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 The frictional force f acting upward and parallel to the inclined plane |
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| Net force acting downward and parallel to the inclined plane = Mg sin q - f. |
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| This net force accelerates the cylinder. |
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| The equation for the linear acceleration (a) of the centre of mass is as follows: |
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| Ma = Mg sin q - f ----------- (1) |
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| If I is the moment of inertia of the cylinder about the centre of mass, then torque acting on the cylinder = I a = RF or |
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| From equation (1), |
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| This is the expression from the linear acceleration of the cylinder. |
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| Let us now calculate the frictional force 'f'. |
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| From equation (1), |
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| where ms is the coefficient of static friction. |
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| The normal reaction may be represented by N or R. |
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