Recent content by Awwnutz

The parallel axis theorem says: I = Icom + Mass*Radius^2 So i have the mass of the disk, and the radius, how do i find the Icom?

for part b its is telling me to use the parallel axis theorem. I'm not too sure how to set it up. I know its the integral of a distance multiplied by the change in mass over a certain limit. But i don't know where to start.

part a for I, i did: Mblock*distance^2 2.6kg*(.26m-.15m)^2 = .03146kg*m^2, and this is correct.

http://img267.imageshack.us/img267/9499/diskwithblockpu2.gif A uniform disk of mass Mdisk = 4.9 kg and radius R = 0.26 m has a small block of mass mblock = 2.6 kg on its rim. The disk rotates about an axis a distance d = 0.15 m from its center that intersects the disk along the radius on...

http://img530.imageshack.us/img530/7629/pnxl1.gif A particle p traveling with a speed of vpi = 3 m/s hits and scatters elastically from another particle N, initially at rest. Particle p is deflected through 90°, leaving with a speed of vpf = 2.7 m/s, and a mass mp = 2 kg. a) What angle...

Yeah i definitely overlooked what i was given. They give me the mass and all i needed to do was find the weight. I start doing all these equations and formula's that i figure everything needs to be some type of formula so i overlook something so simple. I feel like an idiot :) Thanks for...

Alright i figured that one out. I'm stuck on finding the weight of the ball...what equation would i want to do to find that? Should i set the change in Kinetic energy = the work done by gravity to find the mass?

Now I'm on to the time interval between bounces. Would i just use 2-D Kinematics?

the momentum of the ball moving back up would be (.035kg * 20.199m/sec) = .706965J -.707 - .707 = 1.41J total. Ok got it.

The change in momentum? The vector as the ball moves downward would be negative, but as it bounces and begins to move upward it would then be positive.

http://img395.imageshack.us/img395/6158/steelballxu7.gif A 35 g steel ball bounces elastically on a steel plate, always returning to the same maximum height h = 20.8 m. a) With what speed does the ball leave the plate? b) What is the magnitude of the total change in momentum of the...

I figured it out by looking at other questions just like this. My resultant in the x direction was wrong. I just did 40cos(50.625) when i needed to do 40cos(50) - 0.5. The rest was pretty easy after that.

http://img266.imageshack.us/img266/3747/cannonlr0.gif A circus cannon, which has a mass M = 5000 kg, is tilted at θ = 50°. When it shoots a projectile at v0 = 40 m/s with respect to the cannon, the cannon recoils along a horizontal track at vcannon = 0.5 m/s with respect to the ground...

i figured it out Wgrav = m*g*(sin theta)*d Wfric = (Coefficient of friction)*(m*g*cos theta)*d after that it was quite easy to figure out the rest. Thanks PhanthomJay!

ok so your saying Wgrav = -(20kg)(9.81m/s^2)(sin33)(1.9m) And if your turn your coordinate axis so the normal force is going in the positive y direction wouldn't that make the normal force equal and opposite of the weight of the block?