Recent content by uSee2

This is perfect! Thank you so much, this clears up a lot and makes a lot of sense. Also, was this derivation in a textbook? I feel that my textbooks should've mentioned this derivation as this seems very critical to understanding, I might have missed something Also off-topic, but do you have a...

For the first part, it would take the same amount of energy, right? But isn't this different since all parts of the wheel at the edge are moving at the same tangential velocity, while when the cylinder is rolling without slipping the point of contact has a velocity of 0 while the top has twice...

I'm still confused about the idea that there is no rotation about the center of mass. If there was no rotation about the center of mass, wouldn't there be no translational kinetic energy? But it wouldn't move in a circle like in post #11 since the axis of rotation is constantly shifting.

I do see that it is rotating, but since the axis of rotation is changing every moment it is as if it were moving translationally. But if it has only rotational kinetic energy instantaneously for every instant, wouldn't that mean that it only has rotational kinetic energy for the entirety of its...

After reading it, wouldn't that mean that the equation definitely wouldn't work? Since if the axis of rotation was indeed the point of contact, that would mean that there is no translational motion at all. So it would just be: ##KE_{total} = \frac {1} {2} I \omega^2## where ##I## is the moment...

Is the equation just something that works for rigid body motion? I can't exactly find the explanation online somewhere, but do you think that the explanation would involve calculus? If it does, I'm fine with just accepting that the equation "just works" since I still have no knowledge of...

Thank you for answering! I do see that in the case of an object spinning while moving through space, that would be true since the tangential velocity at a distance R away from the center of mass would be constant at any point. But when an object is rolling without slipping, the bottom is at rest...

Given that there is a cylinder rolling without slipping down an incline, the method I was taught to represent the KE of the cylinder was: ##KE_{total} = KE_{translational} + KE_{rotational}## ##KE_{total} = \frac {1} {2} mv_{cm}^2 + \frac1 2 I \omega^2## Where "cm" is the center of mass, and...

Ohh!! I can't believe I made that mistake. Thank you so much for catching that. I see now that ##v_2 = \frac 1 2 v_1##

I do see that they aren't the same. But when I tried it again with different symbols just like @scottdave said: 2 is after the label for when radius is reduced, 1 is the label for before. ##F_{c2} = \frac {mv_2^2} {R_2}## ##F_{c1} = \frac {mv_1^2} {R_1}## ##F_{c1} = F_{c2}## ##R_2 = \frac 1...

But couldn't the denominator decrease and the numerator increase instead and give the same ratio? ##\frac {mv^2} {\frac R 4} * \frac 4 4 = \frac {4mv^2} R## Then it is equal to ##\frac {m(2v)^2} R## Isn't this true?

The answer key states that the new tangential speed is half the original speed. However, this isn't correct right? It should double. My proof: ##F_c = \frac {mv^2} R## ##F_c = F_t## ##\frac {mv^2} {\frac R 4} = \frac {m(2v)^2} R## If centripetal force were to stay constant. As such, tangential...

Thank you so much! And so this is the correct answer rather than the ##\frac {12Mg} {H}## in the answer key?

Ahh I see, I forgot the mass doubles. The corrected speed is ##\frac 1 2 \sqrt{2gH}## After doing it all again, I got ##\frac {8Mg} {H} = k## Is this correct?

I see where I went wrong, I accidently substituted in H/2 instead of H for my first kinematic equation. Speed after they collide is: ##\sqrt{2gH} = v_f## New KE is: ##\frac 1 2 2Mv_f^2 = 2MgH = KE## Doing all my calculations again is: ##\Delta PE_s = KE + PE_g## ##\frac 1 2 MgH + \frac 1 8 kH^2...