Explanation Given: The velocity υ is 8 m/s . The mass of the bob is 2 kg . The radius r is 2 m . Write the equation of centripetal acceleration a n = υ 2 r (I). Here, centripetal acceleration is a , the velocity is υ and the radius is r . Write the expression for change in arc length, d s = r d θ (II). Here radius is r ., change in arc length is d s ad change in angle is d θ . Write the expression for tangential acceleration. a t = υ d υ d s υ d υ = a t d s (III). Here tangential acceleration is a , velocity is υ and rate of change of velocity with respect to distance is d υ d s . Draw the free body diagram of bob at angle θ is 0 ° as shown in Figure (1). Refer Figure (1) write the formula for equation of motion the normal direction. ∑ F n = m a n (IV). Here, summation of total force acting on normal direction is ∑ F n mass is m and acceleration in the normal direction is a n . Refer Figure (1) calculate the normal force in the vertical direction. ∑ F n = T (V). Draw the free body diagram of bob at angle θ is 30 ° as shown in Figure (2). Refer Figure (2) write the formula for equation of motion in the tangential direction. ∑ F t = m a t (VI). Here, summation of total force acting on tangential direction is ∑ F t mass is m and acceleration in the tangential direction is a t . Refer Figure (2) resolve the forces in the horizontal direction. ∑ F t = − m ( 9.81 m/s 2 ) cos θ (VII). Refer Figure (2) write the formula for equation of motion the normal direction. ∑ F n = m a n (VIII). Here, summation of total force acting on normal direction is ∑ F n mass is m and acceleration in the normal direction is a n . Refer Figure (2) calculate the normal force in the vertical direction. ∑ F n = T + m ( 9.81 m/s 2 ) sin θ (IX). Conclusion: When bob at angle θ is 0 ° Substitute 2 m for r and 8 m/s is υ in Equation (I). a n = υ 2 r = ( 8 ) 2 2 Substitute ( 8 ) 2 2 for a n and 2 kg for m in Equation (IV). ∑ F n = 2 ( ( 8 ) 2 2 ) = 64 Substitute 64 for ∑ F n in Equation (V). ∑ F n = T T = 64 N Thus, the initial tension in cord after the bob is released when angle θ is 0 ° is 64 N . When bob at angle θ is 30 ° Substitute 2 kg for m in Equation (VI)

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Chapter 13, Problem 80P

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Chapter 13, Problem 79P

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The initial tension in cord after the bob is released when angles θ are 0 ° and 30 ° .

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