Recent content by StudentOfScience

You bring up an interesting point. One of the main points argued in Lockhart's A Mathematician's Lament (https://www.maa.org/external_archive/devlin/LockhartsLament.pdf) is the inadequacy of mathematics education nowadays. The nature of the essay is from a 'purist' perspective of mathematics...

Thank you for pointing that out! Currently, I have not formally studied Euler angles and tops (and all the other complexities in rigid body motion like the inertia tensor). If you are wondering, I am using Morin and Kleppner as a primary texts (more-so Morin) for self-study. I'm not too far into...

Ah yes, thank you for pointing that out. This makes more sense now; velocity will only be parallel/anti-parallel to ## \hat \theta ## if the position vector is normal to the trajectory. I suppose some of my sketching led me astray. Of course it may happen that at some points along the trajectory...

Bump. Also note a correction: the Newtonian equations of motion should actually read (I forgot to include the m on the side of the accelerations): $$ m(\ddot l -l \dot \theta^2) = mg\cos\theta - \frac{1}{2} \alpha \dot l \sqrt{ \dot l+l^2\dot \theta^2} - k(l(t)-l_{eq}) $$ $$ m(l \ddot \theta...

Homework Statement Consider a point mass of mass m suspended from an ideal, massless spring. Let ##\theta ## be measured from the vertical. Find the displacement of the mass as a function of time if the spring is initially stretched/compressed a distance ## l_0 ## and has an initial velocity...

Ok, I've come to the realization that when resolving a vector in polar coordinates, the base vectors are drawn with respect to the position of the point of interest (the cart in our case) at a certain instant in time. That is, if the cart is at some point on the loop, ## \hat r, \hat \theta ##...

Could you explain? I always thought that if two vectors have the same direction and magnitude, they are equivalent (so it doesn't matter where they are on the euclidean plane - translation doesn't affect the equivalence of the vectors). And here gravity is independent of time, so I am not quite...

Even if gravity points in the same direction (with approximately the same magnitude - the same for our purposes) throughout the cart's entire path?

I'll start with the cartesian form for gravity and the polar form, and then I'll break the polar base vectors into cartesian ones and equate components: $$ \vec F_g = -mg \hat j = F_{gr}\hat r + F_{g\theta} \hat \theta \rightarrow -mg \hat j= F_{gr}\cos\theta \hat i + F_{gr}\sin\theta \hat j -...

Thank you so much! That makes sense now. I think that equation above comes from a mistake that I always make with polar coordinates: I tend to forget about the theta dependence of the polar base vectors, and thus often include an extra theta in there. In other words, $$ \vec F_n = F_{nx} \hat i...

Maybe the misunderstanding is that $$\vec F_n = F_{nx} \hat i + F_{ny} \hat j = -|\vec F_n|\hat r = -|\vec F_n|(\cos\theta(t) \hat i + \sin\theta(t) \hat j) $$ But wouldn't it still be ok to say ## \vec F_n = F_{nr} \hat r + F_{n\theta} \hat \theta ## ?

Looking back at the problem statement, it is essentially a pendulum for ## \theta## varying from 0 to 2pi, except that the 'string' (the radius here) doesn't stretch or anything complicated like that. I've looked at the simple pendulum just a bit right now; I'm not sure why tension is not being...

Here, I am concerned when ## v_0< R\theta_0 \sqrt{\frac{1}{mR}} ##, which makes the square roots negative. Does this correspond to when the cart doesn't have enough speed to make it around the loop? Also, what should I do with the plus-minuses? Leaving them would mean that there are multiple...

Solving that differential equation gives $$ \theta (t) = \sqrt{mR\gamma} \frac{\left (\frac{\theta_0}{\sqrt{mR\gamma}}+\sqrt{1+\frac{\theta_0^2}{mR\gamma}}\right)^2 \exp \left (\frac{\pm 2t}{\sqrt{mR}}\right )-1}{2 \left...

Normally, $$ \vec F_g=-mg \hat j $$ So, $$ -mg \hat j = mg \hat r $$ where the ## \theta## is ## \frac{3\pi}{2} ##, which makes the cosine in the r hat go to 0 ( ##\hat r = \cos\theta \hat i + \sin\theta \hat j ##) Yes, I see that I did make a mistake; thanks for pointing that out. So that...