Explanation Draw the schematic diagram of the system before and after the impact. Figure-(1) Write the expression for the summation of force in y- direction using the Figure (1). ∑ F y = m ( a G ) y (I) Here, the acceleration of the plate about its mass center G is ( a G ) y , the mass of the semicircular plate is m and the sum of the applied force in y- direction is ∑ F y . Write the expression for the force. F = ∑ F y + ∑ F x (II) Here, the summation of the force in the x- direction is zero due the zero acceleration in the respective direction. Substitute 0 for ∑ F x in Equation (II). F = ∑ F y + 0 F = ∑ F y Substitute F for ∑ F y in Equation (I). F = m ( a G ) y ( a G ) y = F m (III) Write the expression for the mass moment of inertia of the semi circular plate about the mass centre G . I G = 1 2 m r 2 − 16 9 π 2 m r 2 = m r 2 ( 1 2 − 16 9 π 2 ) = m r 2 ( 0.5 − 0.18 ) = 0.320 m r 2 (IV) Here, the radius of the semicircular plate is r . Write the expression for the sum of the moments of the semicircular plate about the mass centre G . ∑ M G = I G α (V) Here, the mass moment of the inertia of the semicircular plate about the mass centre G is I G and the angular acceleration is α . Write the for the sum of the moments of the semicircular plate about the mass centre G . ∑ M G = − r F Substitute − r F for ∑ M G and 0.320 m r 2 for I G in Equation (V). − r F = ( 0.320 m r 2 ) α α = − r F 0.320 m r 2 α = − 3.126 F m r (VI) Write the expression for the position vector of point P with respect to point G . r P / G = x P i + ( y p − 4 r 3 π ) j (VII) Write the expression for the vector relation of the acceleration of the mass center G . a G = ( a G ) y j (VIII) Here, the acceleration vector of mass center G is ( a G ) y . Substitute F m for ( a G ) y in Equation (VIII). a G = ( F m ) j (IX) Write the expression for the angular acceleration vector. α = α k (X) Here, the unit vector in z- direction is k . Substitute − 3.126 F m r for α in Equation (X). α = ( − 3.126 F m r ) k (XI) Write the expression for the relation for absolute acceleration of the point P and absolute acceleration of point G . a P = a G + α r P / G − ω 2 r P / G (XII) Here, the initial acceleration of the point P in the plate is a P

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Chapter 6.9, Problem 220RP

To determine

The coordinate of the point P in the plate which has zero initial acceleration.

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Chapter 6.9, Problem 219RP

Chapter 6.9, Problem 221RP

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