Recent content by srmico

I also think it's wrong, but that's what my teacher assistant told me to do. I asked him about the gravity part, and he said that it's inside the "mgh" (I know it's wrong). As I said, this is from last semester, so I will not spend any more time trying to solve it, because I don't have it (I...

I ended up solving it this way, if anyone is interested: From the conservation of energy we have that \begin{equation} \frac{1}{2}mv^{2}=mgh+\Delta E \label{eq:1} \end{equation} where $\Delta E$ is the energy difference due to friction force. \begin{equation*} \Delta E=E_{f}\text{(energy lost...

As I said in my first post, I'm not allowed to solve the exercice with energy. It has to be using torque. Sorry I haven't post anything, but the truth is that i have exams in 10 days, and I won't have time to find a final answer.

I can't do it. Every time I try I find more mistakes. One of the equations I had was ##v_{1}=\frac{v}{1-\frac{a}{r\alpha}}##. But ##a=r\alpha## is another condition for rolling, meaning that I have ##v_{1}=\frac{v}{0}##. I have checked every single sign, the formulae and everything. I must be...

So substituting (3) on (4) we get ##v_{1}-a\frac{v_{1}}{r\alpha}=v## ==> ##v_{1}(1-\frac{a}{r\alpha})=v## ==> ##v_{1}=\frac{v}{(1-\frac{a}{r\alpha})}## And plugging the numbers I got that result. There seems to be a mistake tho. I don't know why but I would expect some g dependence for #v_{1}##...

The expression for ##v_1## comes from ##\omega_{f}-\omega_{i}=\alpha t## for ##\omega_{i}=0## since it's not rolling first and ##\omega_{f}=\frac{v_{1}}{r}## rolling condition. ##v_1## looks good to me since it has the same units as v. Do you see a mistake?

I see what you mean, but I think it still has to be constant so it shouldn't be too important. I can't believe this was 1 problem of 12 in the exam. Not that it's so hard, but the computations are so lengthy. Anyway many thanks, if you notice something wrong let me know :) Ps: I did most of...

Alright, it took me a while (I decided to learn some Latex) but I think I have the answer, or something close. Here is the entire Latex document as I will send it to my professor. The entire mark of the semester depends on this exercice, so feel free to comment And many thanks to Dol Al for the...

This is what I did so far, could you tell me if it is correct? First I will consider the motion from ##t_{0}## until the ball starts rolling at ##t_{1}##. We have that: \begin{gather*} \tau = I\alpha \\ \tau = F_{f}\cdot r = mg\cos(\theta)\mu_{d}\cdot r \end{gather*} Solving for...

I did Fg*sin(z)+Ffriction=ma, solved for a, substituting at x1=x0+vt+1/2at^2, then substituting the t for the t1 that i found and i should get the displacement for the first part?. For the second part, when the ball is rotating I am not sure what to do. I guess now the torque changes sign, but...

I still don't know how to solve the problem. Some help would be much apreciated.

Omg I am mixing up equations. Now should be correct. Anyway I tried doing torque=Friction*Radius=Inertia*alpha. Solving for alpha and substituting at equation w=w0+alpha*t for w=v1/r (rolling condition) i get t1=(2*v1)/(5*g*cos(z)*µ), t1 is time it takes to start rolling. v1 velocity when it...

Sorry it was a typo, i edited it.

Hello, 1. Homework Statement A spherical continuous ball is sliding with a constant velocity v along a frictionless lane. Thereafter it enters an inclined surface (the angle between the surface and the horizontal plane is α) with the coefficient of friction µ between the ball and the surface...

You could use the formula (distance)=(initial velocity)*t + (1/2)*(acceleration)*t^2 to find the time, using initial velocity = 0, acceleration = -9.8 and distance = 4.7. Then you could use the formula (final velocity)-(initial velocity) = (acceleration)*t, using the same acceleration and time...