Please help! Problem: A 0.200-m uniform bar has a mass of 0.720 kg and is released from rest in the vertical position, as the drawing indicates. The spring is initially unstrained and has a spring constant of k = 22.0 N/m. Find the tangential speed with which end A strikes the horizontal surface. (picture is attached).

Okay, so -mgh = ½ mv ² - ½ ks ² + ½ Іω² (I hope).

When I plug it all in (with v being unknown), I get that v=(+/-)4.39254. Then I think I use ω = V /r and r = L / 2 to solve for velocity at point a= w*L. This leads me to the wrong answer. Where am I wrong? Thanks!

Question

Please help! Problem: A 0.200-m uniform bar has a mass of 0.720 kg and is released from rest in the vertical position, as the drawing indicates. The spring is initially unstrained and has a spring constant of k = 22.0 N/m. Find the tangential speed with which end A strikes the horizontal surface.  (picture is attached).

Okay, so -mgh = ½ mv ² - ½ ks ² + ½ Іω² (I hope).

When I plug it all in (with v being unknown), I get that v=(+/-)4.39254.  Then I think I use ω = V /r and r = L / 2 to solve for velocity at point a= w*L.  This leads me to the wrong answer. Where am I wrong?  Thanks!

anonymous · Answer

The pic for the problem is attached.  Please help!  Thanks.

anonymous · Answer

its mgh = ½ mv ² + ½ ks ² + ½ Іω²

anonymous · Answer

then vtangential = v + rw
with r = l/2 and w = v/r

anonymous · Answer

Thanks so much!  But wait.. So I have  mgh = ½ mv ² + ½ ks ² + ½ Іω² ==> (.72)*(9.81)*(.1)=.5(.72)*v^2+.5(22)*(1.24)^2 +.5*((1/3)*(.72)*(.2)^2)*(v^2/(.100^2)), which gives v=(+/-)4.39254i.  Should I have an imaginary number, and if so, what do I do with it?

anonymous · Answer

I'm still befuddled, but thank you anyway, I'm just the confused guy.

anonymous · Answer

Ha!  I got it!  .72*9.81*.1=(1/2)*22*0.1²*(sqrt(5) - 1)²+(1/6)*0.720*0.2²*w², then I found w and plugged it into w=(v/r) and solved.  The answer was 2.12,  thanks so much for giving me a tip!