Applied Maths - Mechanics 62

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