Question:Please first use equation (3) and solve $̇()$
By definition: let $a$ =second derivative of $\theta$
$b$ =first derivative of theta
Here, $ℎ$ is the $y$-coordinate of $$ and we take these parameters: $ℎ= (3/8), _ =(83/320)^2, = 0.5, =250, (=0)= 60$ degree .
$$ a=-hm\sin(\theta)(g+rb^2)/(I_c+m(r^2+h^2-2hr\cos(\theta)))\ \ \ \ (eq3) $$
the definition of eq3 enter image description here
we further simplify the roly-poly toy on the plane, where point $$ is the center of gravity, $$ is the geometric center of the hemisphere, $ℎ$ is the distance between $$ and $$, $$ is the gravity of this hemisphere, $$ is the contacting point, $$ is the radius of the hemisphere, $$ is the variable to describe the movement of this hemisphere, the normal force and the friction from the ground to the hemisphere are respectively $$ and $$.
the graph of roly-poly toy model
attempt:I just simplify the eq3 and put theta(t=0) =60 degree into the eq 3,
then I solve first derivative of theta (t=0)is (surd of 15)/3 and hc(t=0) is 13/16r,but I don't know how to find the value of theta(t),first derivative of theta(t) and hc(t).
plz help me, thx!
Hint.
As
$$ C = A + (h\sin\theta,r-h\cos\theta) $$
we have
$$ \dot C = \dot A + (h\cos\theta,h\sin\theta)\dot\theta = (-r,0)\dot\theta+(h\cos\theta,h\sin\theta)\dot\theta $$
and now
$$ L = \frac 12m\dot C\cdot\dot C+\frac 12I_c\dot\theta^2-m g(r-h\cos\theta) $$
the movement equations are obtained with
$$ \frac{d}{dt}\left(\frac{\partial L}{\partial\dot\theta}\right)-\frac{\partial L}{\partial\theta}=0 $$
giving
$$ \ddot\theta+\frac{h m \sin \theta \left(g+r \dot \theta ^2\right)}{m \left(h^2+r^2\right)-2 h m r \cos \theta +I_c}=0 $$ etc.