Recent content by Warmacblu

Given ∑(1/2)mω²r². Does that mean you solve the problem by setting it equal to Ke and solving for omega?

I understand the concept of a square with a circle of negative mass but I don't really know how to approach the problem. I think I should find the center of mass of the square and then the center of mass for the circle and subtract the two values. The center of mass on the square will be...

Thanks for the help.

I looked up the law of sines and law of cosines and I think the law of cosines would be easier to use: c2 = a2 + b2 – 2abcosC c2 = 1702 + 1702 - 2*170*170*cos(43) c = 124.6104 Does that seem okay.

So let me get this straight. As the Earth spins faster, the value of g (9.8m/s^2) remains the same? And it feels like it's decreasing?

I think it might be a right angle because we are taking the angle measurement from the rest position and if I draw and arrow to the right and connect the hypotenuse (170 cm) then the angle turns out to be 90 degrees on the bottom with a 43 degree angle on top.

Homework Statement A circular hole of diameter 26.4 cm is cut out of a uniform square of sheet metal having sides 52.8 cm. Note - The circle is located at the top right of the square. What is the distance between the center of mass and the center of the square? Homework Equations...

sin43 = x/170 x = sin43 x 170 x = 115.9397

Homework Statement A beetle takes a joy ride on a pendulum. The string supporting the mass of the pendulum is 170 cm long. A) If the beetle rides through a swing of 43 degrees, how far has he traveled along the path of the pendulum? B) What is the displacement experienced by the beetle while...

22 rev / 9 sec = 2.44 rev/sec = 15.36 rad / sec. At the end of the time interval it is at 15 rad/s, so I guess the acceleration is negative.

So the force we feel when spinning at a high angular velocity is the centrifugal force and not any change in gravity?

Ah, okay. I see it now. I am unsure if my answer is in rad/sec2 which the question specifies for. I multiplied 15 rad/s times 9 sec - 138.23 rad all divided by 9 sec. The terms seem correct but the negative is still throwing me off.

Does this mean that less is perceived as weight or the actual value of g, let us say 10 m/s2, actually decreases?

Here is what I got: 2((15*9)-138.23) / 92 = -.0798 Does that mean that the disc has a negative acceleration? Also, I still do not understand how you just dropped the at2: wft - theta = at2 - 1/2at2 wft - theta / t2 = a - 1/2at2 This step is wrong. It should be wft - theta = 1/2*at2

How come? Is it like those theme park rides where you spin in a circle while standing and you feel like you are floating upwards?