- #71
Karol
- 1,380
- 22
$$u=y^2~~\rightarrow~~\dot\theta=\sqrt{u(x)=\frac{2g}{(4\mu^2+1)R}[(1-2\mu^2)\sin(x)+3\mu\cos(x)-3\mu e^{-2\mu x}]}$$
$$F_N=mR\dot\theta^2+mg\sin(\theta),~~T=F_N\mu r$$TSny said:Some typos here?
I have to insert ##u(x)=\dot\theta^2## from post #69 into the above and then use it in:TSny said:You know ##\dot{\theta}(\theta)## and you know ##\alpha(\theta)## explicitly from your post #38 (after correcting the typo)
I believe this is off by a factor of ##r##.Karol said:$$\alpha=\left( \frac{R\dot\theta^2+g\sin\theta}{kr^2} \right)\mu$$
Yes. As you can see, it is a very difficult integral to evaluate.I have to insert ##u(x)=\dot\theta^2## from post #69 into the above and then use it in:
$$\omega(\theta) = \large \int \frac{\alpha(\theta)}{\dot{\theta}(\theta)} \small d\theta$$
$$T=I_c\alpha~~\rightarrow~~\alpha=\left( \frac{R\dot\theta^2+g\sin\theta}{kr} \right)\mu$$TSny said:I believe this is off by a factor of rrr.
But i find:VMP said:After a bit of algebraic manipulation we find this result which we will use in a bit:
$$mR\dot{\theta}=\frac{10}{7}mg\cdot sin(\theta)$$
$$\frac{1}{2}mv_{cm}^2+\frac{1}{2}I_{cm}\dot{\varphi}^2=mgR(1-cos(\theta))\; (1)$$VMP said:From equation (1) you should be able to derive that: ##\ddot{\theta}=\frac{5g}{7R}sin(\theta)##
Yea, I just realized i screwed up the differential equation solution, sorry. The energy equation should also contain the plus sign. I'll try to redeem myself when I get the time.Karol said:@VMP
You start from the bottom of the bowl and the ball deaccelerates. why did you take as positive ##+mR\ddot{\theta}\cdot\hat{\theta}## and also f in:
The second and third R's on the right side of the equation should be r's=the ball's radius, not R=the radius from the center of the bowl to the ball's centerVMP said:From angle ##\theta_{0}## ball is rolling and thus energy is conserved:
$$\frac{1}{2}mR^2\dot{\theta}^2+\frac{1}{5}mR^2\dot\theta^2+mgR(1-sin\theta)=mgR(1-sin\theta_{0})+\frac{1}{2}mR^2\omega_{0}^2+\frac{1}{5}mR^2\omega_{0}^2$$
No, first term is kinetic energy of CoM, second term is rotational kinetic energy. Recall, if the ball is rolling without slipping, then [itex]v_{cm}=R\dot{\theta}=r\dot{\varphi}[/itex].Karol said:The second and third R's on the right side of the equation should be r's=the ball's radius, not R=the radius from the center of the bowl to the ball's center
Karol said:Right. but i get a different expression for ##\mu_s## but it's tedious. i understand you differentiated (3) to get ##\ddot\theta=\frac{1}{2}\cos\theta## and inserted it into
$$\ddot{\theta}+\mu_{k}\,\dot{\theta}^2=\frac{g}{R}(cos\theta-\mu_{k}\sin\theta)$$
You renamed ##\mu_k## to ##\mu_s##, right?
I mean in set of equations (1) ##\mu_k## appears, but in eq' (*) ##\mu_s##. i understand we can interchange between them when we deal with non slip.VMP said:You'll have to be more specific with your last ##\mu## related question.
In system of equations marked by (1) [itex]\mu[/itex] is one function which is defined differently on different intervals.Karol said:I mean in set of equations (1) ##\mu_k## appears, but in eq' (*) ##\mu_s##. i understand we can interchange between them when we deal with non slip.