Explanation Draw the free body diagram as shown in Figure (1). Here the normal reaction is N , mass of the block m , the acceleration due to gravity g , velocity is v , angle is θ , frictional force is F f , radius of curvature is r , angle is θ , and the angular velocity of the block is ω . Apply force equilibrium conditions along normal direction. N − m g sin θ = m v 2 r N = m g sin θ + m v 2 r (I) Here, mass of the block m and velocity is v Apply force equilibrium conditions along tangential direction. m g cos θ − μ k N = m a (II) Here, the coefficient of kinematic friction μ k and the acceleration is a . Equate equation (I) and equation (II). m g cos θ − μ k ( m g sin θ + m μ k v 2 r ) = m a a = g cos θ − μ k g sin θ − μ k v 2 r (III) Apply the relations of kinematics of motion in Equation (III). r θ ˙ d θ ˙ d θ = g cos θ − μ k g sin θ − μ k ( r θ ˙ ) 2 r 1 2 r ( d d θ × ( r 2 θ ˙ 2 ) ) = g cos θ − μ k g sin θ − μ k ( r θ ˙ ) 2 r (IV) Here, angular velocity is θ ˙ and change in angular displacement is d θ . Substitute t for ( r θ ˙ ) 2 in Equation (IV). 1 2 r × d t d θ = g cos θ − μ k g sin θ − μ k t r d t d θ + 2 μ k t = 2 g r ( cos θ − μ k sin θ ) (V) From Equation (V) undetermined coefficients, the solution can be decomposed into homogenous solution ( H = C e λ θ ) and particular solution ( P = A cos θ + B sin θ ) that are combined expression can be written as t = H + P . Here, the homogenous and particular solution results. H = C e − 2 μ k θ (VI) P = 2 g r ( 1 + 4 μ k 2 ) × ( 3 μ k cos θ + sin θ − 2 μ k 2 sin θ ) (VII) Final combined solution of undetermined coefficients written as, t = H + P (VIII) Substitute t for ( r θ ˙ ) 2 , Equation (VI), Equation (VII) in Equation (VIII)

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Chapter 3.5, Problem 93P

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