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A small bead ( A) of mass m =2 kg can slide without friction on a fixed horizontal rod, which is led through a diametric hole across the ball. There is another ball (B) of same mass attached to the first ball by a thin ideal string of length L =1.6 m. Initially the balls are at rest, the thread is horizontally stretched to its length and is passing through the rod. Then the ball B is released with zero initial velocity. Consider the instant when the thread becomes vertical and bead A is about to leave the rod. Their speeds and accelerations at this instant are Va , Vb and aA, aB respectively. The force by rod on A is F and the tension in the string is T at this instant ( take g =10 m/s^2)(A) Va/Vb =1 and aB = 20 m/s^2 (B) Vb = 4 m/s and aB =10 m/s^2 (C) Va = 0 and T =F(D) Va = 4 m/s and F/T = 3/2

A small bead ( A) of mass m =2 kg can slide without friction on a fixed horizontal rod, which is led through a diametric hole across the ball. There is another ball (B) of same mass attached to the first ball by a thin ideal string of length L =1.6 m. Initially the balls are at rest, the thread is horizontally stretched to its length and is passing through the rod. Then the ball B is released with zero initial velocity. Consider the instant when the thread becomes vertical and bead A is about to leave the rod. Their speeds and accelerations at this instant are Va , Vb and aA, aB respectively. The force by rod on A is F and the tension in the string is T at this instant ( take g =10 m/s^2)(A) Va/Vb =1 and aB = 20 m/s^2 (B) Vb = 4 m/s and aB =10 m/s^2 (C) Va = 0 and T =F(D) Va = 4 m/s and F/T = 3/2

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Grade:12th pass

1 Answers

felix
15 Points
3 years ago
ddAV =o and AV= V b= u Using conservation of energy U =( gl)^1/2= 4m / s acceleration of centre of mass ac=ab/2and ab=ac+( u//2)^2 /(l/2)=u^2/l =g Thus T = mg = mgB T = 2mg and F โ€“ 2mg = 2mGb/2 ;F= 3mg

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