Recent content by Daeho Ro

Yes. What do you think about the reason?

For example, the displacement with constant velocity is given by ## d = v t. ## What about the displacement with constant acceleration?

In constant acceleration, the displacement is well known and because we know the displacement, you can get the acceleration. This part is really strange. Why don't you check the relevant equations?

I hope you already got the answer for the problem. The key idea is chain rule. The differentiation of ## x## with respect to ## \theta ## is ## dx/d\theta = \cos\theta ##. Then, the integration will change as $$ \int \dfrac{\sin^2\theta}{\cos^5\theta} dx = \int...

Ok, then now I am clear. The cart with mass ## M ## and cart + weight with mass ## M + m## moved by a fixed certain force ## F## and the displacements are ## d_1 ## and ## d_2 ##, respectively.

I thought the certain force only acted at the initial and the cart moves freely. But, it seems you are right. My poor English sometimes confuse me.

I supposed the cart moves without acceleration, that is constant velocity. Since we know the displacement within unknown time interval ##T##, I can only tell you that the velocity is in a form ## 0.5 m / T ##.

What you means is the constant velocity. Is this problem settle down with constant acceleration or constant velocity?

Yes, I know and you almost reach the final goal. ##dx## have to change as ## d\theta ## because the last integration is in a form ## \int (\cdots) d\theta ##. But as you know, ## dx \neq d\theta ##. What is ## dx / d\theta ##? It's really strong hint about this problem.

Then, what is ## dx ## as a function of ## \theta ##?

In this statement, I can get the velocity of cart and cart + 200grams weight, roughly. But I don't know how to get the acceleration.

See, \int \dfrac{x^2}{(1-x^2)^{5/2}} dx = \int \dfrac{ \sin^2\theta}{(1-\sin^2\theta)^{5/2}} dx = \int \dfrac{\sin^2 \theta}{\cos^5\theta} dx \neq \int \dfrac{\sin^2 \theta}{\cos^5\theta} d\theta. You missed something in the lase step.

## dx ## cannot change directly ## d\theta ##. They have to connected by some function.

Oh, I misunderstand what you want to know. I thought you want to calculate the last thing. When you starts from the beginning, there should be ## dx##, the integral variable. Then, I think you may find what you missed during your calculation.

Then, why do you need the acceleration? I know you want to use the Newton's second law, but there are many missing parameters to use this.