I believe you are referring to column vector in your expression and not combination. I like your proof and everything looks right over there. How do i get the tangential acceleration using your method?
I have been trying it both ways using rectangular and polar coordinates. The problem with...
I am not sure about that. Let's say we have a scenario that there is no gravity and that there is constant angular acceleration. How do you prove that the centripetal acceleration is: ##a_c = \frac{v^2}{r}##? For example, let's say that the motion path formula is:
##x = rcos(\theta_0 + \omega_0...
Yes but the centripetal acceleration of ##a_c = \frac{v^2}{r}## is for constant tangential velocity and no tangential acceleration. In the case of the pendulum, there is tangential acceleration.
I guess, zero tangential acceleration would be a non-zero centripetal acceleration. I mean, i know it is weird that there is no centripetal acceleration and i am with you.
Edit: Maybe we should a similar example with a pendulum. The only difference is that there is going to be tension instead...
By doing it your way, if you look only at the centripetal forces, the sum of all the centripetal forces must equal to zero. That is because the resultant does not move into or away from the distance of the stick. So using your formula:
##mgsin\theta + F_{stick} = 0##
##F_{rm stick} =...
Actually, if you review my opening post, ax comes from the x-component of the centripetal acceleration ac. The centripetal force AB is the mass times the centripetal acceleration ac.
The reason why i put the angular acceleration relationship with the tangent acceleration at is because it is...
I have tried various things and they don't seem to come out quite right. Maybe you guys can help me spot any mistakes.\alpha=\frac{a_t}{r}=\frac{gcos\theta}{r}
If:x=rcos\thetaa_x=gcos\theta sin\thetaSo:rcos\theta=\int\int gcos\theta\sin\theta...
Ok. Thanks to your reply TSny, i think i know what mfb meant by the problem statement being incomplete.
So, because of the problem setup we can agree that the motion path of the ball mass will be of circular motion of the form:x=rcos\thetay=rsin\thetaWhere θ is a function of time t.So, if the...
@mfb
That is the full problem statement because I made it up. I want to mathematically show that the path taken by the ball mass at the end of the pinned stick is:x=\sqrt{2}cos\thetay=\sqrt{2}sin\thetaIs there any unknown or missing variable you want to add to the problem?
It is true that i can...
@mfb
When you are talking about the energy equations, i believe you are referring to:\Delta KE = 0\frac{1}{2}I\omega _f^2 - \frac{1}{2}I\omega _i^2= 0\frac{1}{2}I(\omega _f^2 - \omega _i^2)= 0
However, the problem is finding the angular velocity:\omega = \int \alpha dt\omega = \int...
Homework Statement
A massless stick is pinned at point A. And it has a concentrated ball mass of 1kg attached to it at point B.
Given the initial values:
Initial angular position θ=45◦
Initial angular velocity ω=0
Initial angular acceleration α=0
Show that because of the setup of the...
Homework Statement
Let's say that i have a unknown weak acid.
I prepare 10mL of the weak acid diluted in 25mL of water.
I titrate the solution with NaOH. 7.37mL of NaOH was needed for a full titration.
Therefore, half-titration of this acid was 3.69mL.
I measure the pH of the solution with a...
@Zondrina
Well, the thing is that the more i look back when i review my basics, the more i question them. I even ask myself how come 1+1=2? I know it is something very intuitive but i know there is more into it that just that. I also know that there is a proof on that but i was never able to...
@Ray Vickson
Things that are not achievable today doesn't mean they aren't achievable tomorrow.
Things that seem to be correct today may be proven wrong tomorrow.
I am not rejecting numerical solutions, i just don't like it because it is not elegant.
It is just a personal opinion. However, i do...
@micromass
The question becomes how do you know that for all the set of infinity operations and infinity equations, there is a subset of equations which becomes unsolvable no matter the operations used?
@D H
Numerical solutions are great but what i hate about numerical solutions is that they...