Recent content by bkw2694

Homework Statement I've uploaded an image of the problem below. The problem states: Determine the value of P which will cause the homogeneous cylinder to begin to roll up out of its rectangular recess.The mass of the cylinder is m, the mass of the cart is M. Friction of the cart wheels i...

I see that that definitely works mathematically, but I don't understand why it works by looking at the problem. Why is it .27 instead of .18? I don't understand why the distance is .27 when the velocity vector is .18 m away. Thanks again!

Ahhhh, I see now. So I just plugged V_O = .18w into the equation and solved for w first. Thanks so much!

I thought <w>x<OA> was the speed of A relative to O? I thought the equation was sufficient for finding the actual speed of A? How would I solve for thespeed relative to earth?

Homework Statement As shown in the picture, two pulleys are riveted together with a known velocity. I need to find the velocity of two other points. Homework Equations V_A = VB + <ω> x <BA> V = ωr The Attempt at a Solution I've honestly been trying to figure this one out for over a...

Sorry, I one of my other HW problems had t = 4 so I screwed that up. Okay, I figured out my problem. Aside from using the wrong time, I was also using the wrong v. I was using v = 80cosθ, but v was obviously 80 ft/sec and provided as the initial velocity. The incorrect v value with cosθ was...

I must be solving for v (when t = 4) incorrectly. a_n = v^2/p, but when I try doing v = 80cos(theta), then plug gcos(theta) = (80cos(theta))^2/142.2 I end up with theta = 44.3 degrees, but I know the correct answer should be more like 11 degrees.

^whoops, I mixed those two up, at = -gsin(Θ); an = gcos(Θ)

Okay, so I used trigonometry to find at = gcos(Θ). And I suppose I can use the same thing to find an = -gsin(Θ). I tried plugging those into a = sqrt(an^2 + at^2), but I think that causes the Θ to disappear due to sin^2 + cos^2 = 1.

Okay, so I can see that the angle will be getting smaller, but I'm not sure what equation to use to figure this one out. Is this just a projectile equation I'm overlooking or is there calculus involved here? Since I knew the answer from the book I worked backwards and found an, then solved to...

Yeah, sorry, I just now realized what you meant and used the normal projectile equations to find ρ. By using the correct formula, I ended up getting ρ at t=1 is 142.19 ft. So now the only problem I'm having is I can't seem to work out the correct at. Am I wrong in assuming that the magnitude...

So if it isn't following a circular arc, this means that ρ is changing, correct? Or is it that Θ is changing and causing ρ to change? Also, I see the radius of curvature equation is: [1+(dy/dx)^2]^(3/2) / [(d^2y/dx^2)]. But I don't understand how to make that apply to this particular question...

Homework Statement A football player releases a ball at 35° with initial velocity of 80 ft/sec. Determine the radius of curvature of the trajectory at times t = 1 sec and t = 2 sec, where t = 0 is the time of release from the quarterback's hand. For each case, compute the time rate of change...

So for me to integrate the acceleration, I need to know a(s)?

Sorry if I'm not following correctly, but are you saying no integral is required for this question? Solving dv/ds gives me (15/2)s^(1/2). Then multiplying that by the original "v" gives me 5*15/2 (s^2) which after plugging in s = 2 gives me 150mm, the correct answer. Is that the correct setup...