Recent content by ViewtifulBeau

i did that and i got .166 meters. umg = .873*9.81*.468 = 4.008 N F(spring) = kx 4.008 = 24.1 * x x = .166 meters. and this is not right. What am i doing wrong?

Two blocks (m = 0.468 kg and M = 2.41 kg) and a spring (k = 24.1 N/m) are arranged on a horizontal, frictionless surface. Block m is situated on top of block M. (Spring attached to block M) The coefficient of static friction between the two blocks is 0.873. What is the maximum possible...

Romeo takes a uniform 10-m ladder and leans it against the smooth wall of the Capulet residence. The ladder's mass is 30kg, and the bottom rests on the ground 2.55m from the wall. When Romeo, whose mass is 70 kg, gets 88 percent of the way to the top, the ladder begins to slip. What is the...

n/m i got it

w(t) = .30782 = (velocity/circumference) * 2*pi I(t) = .65043 = m(t) * radius^2 I(system) = 12.15315 = (m(w)+m(t)) * (radius)^2 w(system) = w(t)*I(t) / I(system) = .0165 not right! please help.

A toy train track is mounted on a large wheel that is free to turn with negligible friction about a vertical axis as shown in the figure above. The mass of the wheel plus track is 3.59 kg and the radius is 1.79 m. Ignore the mass of the spokes and hub. A toy train of mass 0.203 kg is placed on...

A uniform thin rod of length 0.471 m and mass 4.0 kg can rotate freely on a frictionless horizontal plane about a vertical axis through its center. The rod is at rest when a 3.0 g bullet traveling in the horizontal plane of the rod is fired into one end of the rod. As viewed from above, the...

i did this and found omega to be 8 rad/s and I to be 1.8 then i found the KE to be 57.6 this is now right either.

i still can not figure this one out.

A homogeneous cylinder of radius 30cm and mass 40kg is rolling without slipping along a horizontal floor at 2.4m/s. How much work is needed to stop the cylinder? work = delta_KE in this case so oi figured .5 *40 * 2.4^2 = 115 J but this is not right, what should i do with the cylinder?

well thanks for the quick relpies, ummm yeah here in my book it is I = I(cm) +Mh^2 is I(cm) = 1/12 * M *L^2? M is total mass and h would be... distance from the center of mass?

vf^2 = vo^2+ 2*a*(xi-xf) Vo = 10 each time a = 9.81 each time xi = 100 each time xf = height above ground this should work

A uniform thin rod has a length 3.14 m and mass 2.11 kg. Find the moment of inertia in kg-m2 about an axis that is perpendicular to the rod and passes through the rod at a distance of 0.67 m from the end of the rod. I know that if the axis was in the center of the rod the equation would be...

A 6.59 kg block is released from A on a frictionless track, that slopes down until it reaches the floor, then loops in a circle(kinda like the hotwheels loops). Determine the angular acceleration of the block, alpha, about O, the axis of the loop, at the point P, which is located 45 degrees...

you got it