Recent content by srnj222

if r is the position it would have landed at

Again not sure but here's some progress I made on the first: mv = (m+M)vf mv x (3L/4) = Ix I = 1/3 ML^2 + m (3L/4)^2 I = (1/3 M + 9/16 m)L^2 vf (center of mass) = mv./(m+M) w = (3/4 mv.)/(1/3 M + 9/16 m)

I am currently an undergraduate at dartmouth college, in love with both medicine and engineering, and I've been looking for a way to do both. A friend mentioned he had heard about combined M.D./Ph/D programs in biomedical engineering, which sound awesome, but everywhere i look, I can only find...

I think I've oslved the second one now, was off by a lot before, let me know, also, still lost on the first: 2: mgh = 1/2 I w^2 mg(L/2)=1/2 (1/3mL^2)w^2 g=(1/3)Lw^2 w^2=3g/L w=sqrt(3g/L)=5.422 radians/second v = w * r = 5.422 * 1m = 5.422 m/s

V and v are different, one is velocity of the bullet, the big V is that of the bob. yes, you have 2 equations and 2 unknowns (v and V) so solve one and substitute as you suggested

The center of mass of the shell will continue on the initial trajectory: cm = ((1/5)m*0 + (4/5)m*d)m cm = (4/5)d d = (5/4)cm where cm is the position it owuld have landed at

Since it is the minimum speed at which it will make it over, you calculate v such that the kinetic energy at the top is zero (no velocity) then: mv = m(v/2) + MV (conservation of momentum) 1/2MV^2 = Mg(2l) that should be enough to solve it.

I'm having trouble with these two "classic meter stick problems" 1: A rod of length L and mass M stands vertically on a flat frictionless surface. A wad of putty of mass m and initial velocity v strikes the stick at a right angle at height 3/4 L. The collision is perfectly inelastic. Find the...