A ball on slope
球がスロープを「転がり落ちる」のと「滑り落ちる」のでは、挙動が異なります。
「転がり落ちる」場合、「回転による慣性モーメント」を考慮します。
# a ball on slope
- $ \theta $ : the angle between slope and horizon
- $ H $ : initial Height
- $ r $ : radius of a ball
- $ M $ : Mass (質量)
- $ I $ : inertia moment (慣性モーメント)
- $ y $ : current height
- $ s $ : current length progressed (進んだ距離)
- $ v $ : verocity
- $ a $ : acceralate
- $ \omega $ : angular speed (角速度)
Conservation of energy;
$$ MgH = \frac{1}{2} M v ^ {2} + \frac{1}{2} I \omega ^ {2} + Mgy $$
Inertia moment of a ball is
$$ I = \frac{2}{5} M r ^ {2} $$
So,
$$ MgH = \frac{1}{2} M v ^ {2} + \frac{1}{5} M r ^ {2} \omega ^ {2} + Mgy $$
Because there are no slips, it follows;
$$ v = r \omega $$
So,
$$ MgH = \frac{7}{10} M v ^ {2} + Mgy $$
and differentiate this equation by time $t$
$$ 0 = \frac{7}{5} M v a + M g \dot{y} $$
$s$ is represented by $ \theta $ and height
$$ s \sin{\theta} = H - y $$
$$ \sin{\theta} = - \dot{y} $$
Hence we get;
$$ Mgv \sin{\theta} = \frac{5}{7} M v a $$
$$ a = \frac{5}{7} g \sin{\theta} $$