Recent content by jdboucher

Yes you're right. Silly me. I got it now. Thank you for your help.

I mean that I assumed the bullet was a point and used mr^2 to solve for its moment of inertia. Thats wrong obviously

(1/12)(6.55+1.49)(2.29)^2 + (1.49)(2.29/2)^2 = 6.12 Its wrong again. Am I not supposed to use the parallel axis theorem?

Ifinal = (1/12)ML^2 = (1/12)(1.49 + 6.55)(2.29)^2 Ifinal = 3.51 Is that wrong?

I figured it out thanks

I assume the angle is 90. So L = mvrsin90. If I do that, set it equal to Iw, I get that w= 9.52. According to the online site I'm using that's wrong. Is it my angle?

So then I need to determine r. Is that d/2?

X: Fwmanx + Fwladderx = Fwall Y: Fwmany + Fwladderx = Fn Torque: I'm confused on how to determine clockwise or Counterclockwise Torque.

0.5(1.49)(19.6)^2 = 0.5(1/12)(1.49+6.55)(2.29)^2w^2 I solved this and got w = 12.76 The answer is wrong though. I checked my work. Did I do something wrong?

so can I do this: 0.5mv^2 = Iw

Is it kinetic to potential?

Does translational (i think that's what its called) kinetic energy relate to rotational kinetic energy?

Homework Statement A 9.8 m ladder whose weight is 257 N is placed against a smooth vertical wall. A person whose weight is 464 N stands on the ladder a distance 1.9 m up the ladder. The foot of the ladder rests on the floor 7.056 m from the wall. Calculate the force exerted by the wall...

Homework Statement A projectile of mass m = 1.49 kg moves to the right with speed v0 = 19.6 m/s. The pro- jectile strikes and sticks to the end of a sta- tionary rod of mass M = 6.55 kg and length d = 2.29 m that is pivoted about a frictionless axle through its center. Find the angular...