標題:
Applied Maths - Mechanics 62
發問:
A uniform rod AB of mass 2m kg ang length 2a m is suspended in equilibrium from a smooth horizontal axis through A. A particle of mass m kg strikes the rod at its centre with a velocity of 4u ms^-1 at 30 degree to the vertical. If the particle is smooth and inelastic, find the angular velocity of the rod after the... 顯示更多 A uniform rod AB of mass 2m kg ang length 2a m is suspended in equilibrium from a smooth horizontal axis through A. A particle of mass m kg strikes the rod at its centre with a velocity of 4u ms^-1 at 30 degree to the vertical. If the particle is smooth and inelastic, find the angular velocity of the rod after the collision and the impulsive action between the rod and the particle.
最佳解答:
Moment of inertia of the rod AB about a horizontal axis through A, I = 1/3 (2m)a2 + 2ma2 = 8ma2/3 (// axis theorem) After the particle sticks on the rod (since the collision is inelastic), then the moment of inertia of the system, I' = 8ma2/3 + ma2 = 11ma2/3 We just need to consider the velocity of the particle perpendicular to the rod. As there is no external torque in the system, by conservation of angular momentum, Initial angular momentum = Final angular momentum ma(4ucos30*) = I'w 2mau sqrt3 = 11ma2/3 w Angular velocity after collision, w = (6sqrt3)u / 11a The linear velocity of the particle after collision, v = aw = (6sqrt3)u / 11 So, the impulsive action = mv - m(4ucos30*) = mu[(6sqrt3)/11 - 2sqrt3] = -(16sqrt3)mu / 11 2009-08-24 09:17:10 補充： Impulsive action 已out of syllabus 2009-08-24 09:24:08 補充： 本書的答案沒有sqrt3，應該是它錯，因為它沒有考慮分開component來計。所以是不對的。 2009-08-24 09:25:16 補充： 我忘了寫單位，因為你的題目是有給單位的。 應該是： Angular velocity after collision, w = (6sqrt3)u / 11a rads^-1 The impulsive action = -(16sqrt3)mu / 11 Ns 只考慮magnitude，便得到 (16sqrt3)mu / 11 Ns
其他解答:
Applied Maths - Mechanics 62
發問:
A uniform rod AB of mass 2m kg ang length 2a m is suspended in equilibrium from a smooth horizontal axis through A. A particle of mass m kg strikes the rod at its centre with a velocity of 4u ms^-1 at 30 degree to the vertical. If the particle is smooth and inelastic, find the angular velocity of the rod after the... 顯示更多 A uniform rod AB of mass 2m kg ang length 2a m is suspended in equilibrium from a smooth horizontal axis through A. A particle of mass m kg strikes the rod at its centre with a velocity of 4u ms^-1 at 30 degree to the vertical. If the particle is smooth and inelastic, find the angular velocity of the rod after the collision and the impulsive action between the rod and the particle.
最佳解答:
Moment of inertia of the rod AB about a horizontal axis through A, I = 1/3 (2m)a2 + 2ma2 = 8ma2/3 (// axis theorem) After the particle sticks on the rod (since the collision is inelastic), then the moment of inertia of the system, I' = 8ma2/3 + ma2 = 11ma2/3 We just need to consider the velocity of the particle perpendicular to the rod. As there is no external torque in the system, by conservation of angular momentum, Initial angular momentum = Final angular momentum ma(4ucos30*) = I'w 2mau sqrt3 = 11ma2/3 w Angular velocity after collision, w = (6sqrt3)u / 11a The linear velocity of the particle after collision, v = aw = (6sqrt3)u / 11 So, the impulsive action = mv - m(4ucos30*) = mu[(6sqrt3)/11 - 2sqrt3] = -(16sqrt3)mu / 11 2009-08-24 09:17:10 補充： Impulsive action 已out of syllabus 2009-08-24 09:24:08 補充： 本書的答案沒有sqrt3，應該是它錯，因為它沒有考慮分開component來計。所以是不對的。 2009-08-24 09:25:16 補充： 我忘了寫單位，因為你的題目是有給單位的。 應該是： Angular velocity after collision, w = (6sqrt3)u / 11a rads^-1 The impulsive action = -(16sqrt3)mu / 11 Ns 只考慮magnitude，便得到 (16sqrt3)mu / 11 Ns
其他解答:
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