A thin uniform stick of length 4m and mass 3 kg is hung vertically from a support 1m from its end. The stick is initially at rest. A point mass of 7 kg travelling at 6.11 m/s is moving and impacts the stick at the bottom-most point. Determine how high the stick and point mass will rotate before coming to rest momentarily.(adsbygoogle = window.adsbygoogle || []).push({});

I applied the conservation of angular momentum theorum

L = r (moment arm) mv

= (3)(7)(6.11) = 128.31

I = (summation) m r(moment arm)^2 = 3 (1)^2 + 3 (3)^2 = 30 kg m^2

(where 1 and 3 are the distances from the pivot point to the center of mass and pivot point to the point of impact, respectively)

I` = I(rod) + 7(3)^2 = 93 Kg m^2

L = I omega

128.31 = 93 omega

omega = 1.38 rad/sec

Kinetic energy is then transferred to potenial:

1/2 I` omega ` ^2 = mgh

1/2 (93) (1.38) ^2 = mgh

88.51 = g[7 (3cos theta) + 3 cos theta]

9.022 = 24 cos theta

theta = 67.92 deg.

Are the moment arms that I used for the (I = (summation) m r(moment arm)^2) formula correct? Assuming I did my calculations correct, how do I determine the height from the information that I got?

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# Rod rotates when hit by point mass/ Cons. of angular momentum

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