Recent content by annamal

So you think the way I have defined ##\theta## is clockwise b/c we are measuring the angle from the horizontal to the pendulum string correct? How come we cannot measure the angle from the pendulum string to the horizontal and say that ##\theta## is defined counterclockwise (the way I have the...

I think you meant to say ##\vec i## has to be pointing to the right and ##\vec j## has to be pointing up. That is interesting that the ##\frac{dn_1}{dt}## and ##\frac{dn_2}{dt}## ended up that way. In order for my way of calculating them to be true, we would have to be rotating clockwise so...

For this ##\dot{\vec n_1}## and ##\dot{\vec n_2}## I have them negated. ##\dot{\vec n_1} = \dot{\theta}\vec k \times \vec n_1 = \dot{\theta}\vec n_2## where ##\vec k## is pointing out of the screen and ##\vec n_1 \times \vec n_2 = \vec k##. ##\dot{\vec n_2} = \dot{\theta}\vec k \times \vec n_2 =...

$$\vec a = \frac{d(-v sin\theta \vec n_1 + v cos\theta \vec n_2)}{dt} = (-asin\theta - 2v cos\theta \dot{\theta})\vec n_1 + (-2vsin\theta \dot{\theta} + acos\theta)\vec n_2\ (2)$$ $$\vec a \neq -asin\theta \vec n_1 + acos\theta \vec n_2\ (1) \neq (2)$$

I forgot to add the negative sign to ##vsin\theta## but everything else is the same

Above I forgot to add a negative to ##\vec n_1## so ##\vec v = -v \vec i = -v sin\theta \vec n_1 + v cos\theta \vec n_2## not just ##v\vec n_1##. But the equations (1) and (2) should be correct.

I already included and simplified their derivatives in equation (2)

Yes ##\vec n_2## points along my pendulum rod as it moves and ##\vec n_1## points perpendicular to that constantly.

##\vec n_2## points to the diagonal top left. The velocity v is a function of t. So for example ##2t^2## and a = dv/dt. Putting the velocity vector into ##\vec n_1## and ##\vec n_2## terms. $$\vec v = v sin\theta \vec n_1 + v cos\theta \vec n_2$$ $$\vec v = -v \vec i$$ $$\vec a = \frac{d\vec...

Ok, so the contact point on the ground is moving? Like the contact point is sliding with respect to the ground?

I think the problem is not saying that the contact point has an instantaneous velocity of 0 when it says the contact point traverses with v0. I am confused about why the problem has to say the contact point traverses with v0 and not just say the wheel traverses the hill without slipping if the...

Resolved, that was my mistake. The distance is d - h/2.

For this translating block problem, below is the solution. I was wondering why if I took the moment about the center of gravity G, the answer for F would no longer be the same because ##I_G \alpha = -\mu_k N (h/2) + N (b/2) - F*d = 0## because ##\alpha = 0## $$F = \frac{-\mu_k mg (h/2) +...

What is this "lab" you are referring to?

ok but how do you explain this mathematically? I thought ##\frac{d\vec r}{dt} = \vec v^{P/C}##. Since ##\vec r^{P/C} = (R/2)*(5 + 4 cos\theta)^{1/2} \vec e_r##, the derivative of that is ##(R/4)*(5 + 4cos\theta)(-4sin\theta)^{-1/2}\dot{\theta} \vec e_r + ...\vec e_{\theta}##. Mathematically like...