Recent content by kitz

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That makes sense. How would I go about finding the moment of inertia about the nail? ...This is a longshot, but would the moment of inertia be 2MR^2?? (using the parallel-axis theorem...)

Hello! I did this problem and have gotten it wrong-- but my math is right. Maybe I'm missing something? Here's the question: We want to support a thin hoop by a horizontal nail and have the hoop make one complete small-angle oscillation each 2.0 s. What must the hoop's radius be? So...

So here are the speeds at both: s_{e} = \sqrt{\displaystyle{\frac{2M_{e}G}{R_{e}}}} s_{\alpha}=\sqrt{\displaystyle{\frac{2M_{e}G}{R_{e}\alpha}}} Now, I have to express s_{\alpha} in terms of s_{e} and \alpha ...You mentioned that I should take a ratio of the two? But when I do that, I...

AHH... for the first question, potential energy part, it was in km, and I was treating the kilometers as meters... However, for the first part of the question, the speed is in km/h. Should I convert to m/h or m/s? For the KE formula, does it have to be in m/s?

Hmm... In using the given speed and plugging it into K = \displaystyle{\frac{1}{2}}mv^2. And in using that, for KE, I'm getting 4.53\times 10^{10}, which isn't correct... Am I doing the right thing? For the Potential Energy, using the distance, I'm getting -2.71\times 10^{10}... Which also...

I have two questions I'm a bit confused on... 1st: Ten days after it was launched toward Mars in December 1998, the Mars Climate Orbiter spacecraft (mass 629 kg) was 2.87 \times 10^{6} km from the Earth and traveling at 1.20 \times 10^{4} km/h relative to the earth. a.) At this time...

Thank you very much! After plugging the numbers in, I got 2.814 Thank you, sir!

would that just be: m_{1}gh = \displaystyle{\frac{1}{2}}m_{1}v^2 + \displaystyle{\frac{1}{2}}I\omega^2 + m_{2}gh+\displaystyle{\frac{1}{2}}m_{2}v^2

Thank you sir! Isn't adding 1/2I*omega^2 relating the rotational and the translational speeds? If not, how do I relate tralsational and rotational to the PE of the 2-kg mass? Thank you! Here it is better written: m_{1}gh = \displaystyle{\frac{1}{2}}m_{1}v^2 +...

Hi, I'm having a bit of trouble finding the right equation for this: The pulley in the figure has radius 0.160 m and a moment of inertia 0.480. The rope does not slip on the pulley rim. Use energy methods to calculate the speed of the 4.00-kg block just before it strikes the floor. To...