I got accepted to a M.S. in materials science program with a B.S in Physics and another B.S. in Math with comparable GPAs both over all and for in major course work. You've actually got a leg up on me in that I never did get into an REU program nor did I get any undergraduate research experience...
If you really want to get down to it the elastic limit is also known as http://en.wikipedia.org/wiki/Yield_(engineering)" and is measured in terms of pressure. For concrete its about 68 MPa (megapascals), for steel ball bearings it's probably about 500 MPa and based just on what golf balls are...
Yup and that generalization holds for all materials. The CR of any material will naturally change a little bit depending on the forces involved. By far the greatest change occurs when you pass the elastic limit. After that any deformation is essentially permanent. The steel works well for the...
http://hypertextbook.com/facts/2006/restitution.shtml" I figured with an almost one meter drop the forces involved would be well below the forces in a golf swing. Obviously the more force you apply to an object the closer you get to its elastic limit. Going beyond the elastic limit of a...
What you are looking at is nearly an elastic collision so you need both of these equations:
m_{1}\ast v_{1i} + m_{2}\ast v_{2i} = m_{1}\ast v_{1f} + m_{2}\ast v_{2f}
\frac{1}{2}m_{1}\ast v_{1i}^{2} + \frac{1}{2}m_{2}\ast v_{2i}^{2} = \frac{1}{2}m_{1}\ast v_{1f}^{2} + \frac{1}{2}m_{2}\ast...
That is correct.
This you messed up a little.
cos(ѳ/2) = u*cos(ѳ)/v' because cos of the angle is adjacent over hypotenuse
So:
v' = u*cos(ѳ)/cos(ѳ/2)
Now plug that into mg cos(ѳ/2) = mv'2/r and solve for r and you should get your answer.
You've got the initial vx from the set up. Assuming we are neglecting air resistance the initial vx is the vx through the entire problem. So when the velocity makes an angle of ѳ/2 with the horizontal you have two parts of the puzzle. You have the horizontal component and the angle that makes...
That's roughly what I meant. Obviously the simplest form is going to take advantage of symmetry and that was addressed as well. But along side of that I was also taught very specifically the motivation for choosing the surface such the E perpendicular to dA was a constant (which includes 0 or...
I'm just wondering when you were taught Gauss's Law what was the justification given for choosing one Gaussian surface over another? When I was took E&M one of the first things explained when I learned Gauss's Law was that you choose a Gaussian surface specifically to keep the component of E...
Look again at https://www.physicsforums.com/showpost.php?p=2762548&postcount=4" For ANY wheel that is rolling without slipping ANY increase in angular velocity generated by ANY tangential force produces a linear acceleration on the center of mass of that wheel. If you wish the wheel to be...
Once your force exceeds that static friction between the tire and the surface you will get slippage. This is not the same as overcoming the rolling resistance of the tire. There is quite a bit of extra force that can be exerted between overcoming rolling resistance and causing tire slippage...
You will note that I started with a constant angular acceleration. This angular acceleration was completely general. That is any angular acceleration will behave in this manner. You will note that the bottom of a wheel is at zero velocity relative to what ever it is rolling on. Thus static...
Don't confuse the direction of the torque with the direction of a force. The direction of a force determines many things. The direction of a torque does nothing but keep track of the sign information. It tells you if the force causing the torque is creating a positive of negative angular...
Start with a constant angular acceleration:
d(d(θ)/dt)/dt = α (constant angular acceleration)
d(θ)/dt = αt + ω_o
θ = (1/2)α(t^2) + ω_o(t) + θ_o
Now for a rolling wheel with θ measured in radians we know θ*r = x or distance traveled by its center of mass relative to the surface it is...