Rolling Down The Stairs

Ok, we shall use the diagram below to analyze the problem ballmovign Where to begin

Since the cylinder is always in contact with the staircase, through out its motion we will consider the motion of the object pivoted through the edge of the stair.

We have from diagram $1$ and $2$ , the moment of inertia through the center of mass, and through the pivot point $O$ . ${ I }_{ cm }=\frac { 1 }{ 2 } m{ r }^{ 2 }\\ I_{ O }=\frac { 1 }{ 2 } m{ r }^{ 2 }+m{ r }^{ 2 }=\frac { 3 }{ 2 } m{ r }^{ 2 }$

Also, as the object is moving it has a degree of freedom $-\frac { \theta }{ 2 } and\frac { \theta }{ 2 }$ , measured from the ${ 45 }^{ 0 }$ mark, as seen from $3$ . The angle $\theta$ sweeped out as the object moves from $1$ to $3$ is given by $2{ sin }^{ -1 }\left( \frac { h }{ \sqrt { 2 } r } \right)$ .

Looking, at diagram $3$ we see that the objects center of mass velocity changes from $v_{f}$ to $v_{A}$ , as it collides with the edge of the stair. This changes is caused by the impulsive forces, at the respective contact points. If we choose our coordinate system, to be at the second contact point, we can avoid doing any calculations which require the change in momentum due to impulsive forces.

So we have

$\quad \quad \quad \quad { L }_{ i }={ L }_{ f }\\ \Rightarrow I_{ cm }{ \omega }_{ f }+m{ v }_{ f }rcos\theta ={ I }_{ 0 }{ \omega }_{ A }\quad \longrightarrow v=\omega r\\$

Plugging in the values for the moment of inertia we have,

$\quad { v }_{ A }=\left( \frac { 2 }{ 6 } +\frac { 2 }{ 3 } cos\theta \right) { v }_{ f }$

Using conservation of energy during the period of rotation.

$\frac { 1 }{ 2 } I{ \omega _{ i } }^{ 2 }+mgrsin\left( \frac { \pi }{ 4 } +\frac { \theta }{ 2 } \right) =\frac { 1 }{ 2 } I{ { \omega }_{ f } }^{ 2 }+mgrsin\left( \frac { \pi }{ 4 } -\frac { \theta }{ 2 } \right) \\ \Rightarrow \frac { 3 }{ 4 } ({ { v }_{ f } }^{ 2 }-{ { v }_{ i } }^{ 2 })=gr\left( sin\left( \frac { \pi }{ 4 } +\frac { \theta }{ 2 } \right) -sin\left( \frac { \pi }{ 4 } -\frac { \theta }{ 2 } \right) \right)$

The object has reached terminal velocity when ${ v }_{ i }={ v }_{ A }!!$

since $\Delta v\ll 1$ , we have ${ v }_{ avg }\approx \frac { { v }_{ f }+{ v }_{ i } }{ 2 }$

Its straightforward from then.

Let's analyze what happens when the cylinder collides with the step's edge by using diagram below. Because the collision is inelastic, the velocity component which passes through the step's edge is reduced to zero, leaving us with only the perpendicular component which can be written as

$v_{f}=v_{i}cos(\theta)$

Cosine law gives

$(h\sqrt{2})^{2}=R^{2}+R^{2}-2R^{2}cos(\theta)\\cos(\theta)=1-\frac{h^{2}}{R^{2}}$

Substituting $cos(\theta)$ into $v_{f}$ equation, we get

$v_{f}=v_{i}(1-\frac{h^{2}}{R^{2}})$

Conservation of mechanical energy (before the cylinder rolls down and right before the collision occurs)

$mgh+\frac{1}{2}mv_{0}^{2}=\frac{1}{2}mv_{i}^{2}\\v_{i}=\sqrt{v_{0}^{2}+2gh}$

The cylinder is in steady-state condition (having terminal velocity), means that velocity after the collision and before it rolls down are the same.

$v_{f}=v_{0}$

Substituting all the equations, we have

$\sqrt{v_{0}^{2}+2gh}(1-\frac{h^{2}}{R^{2}})=v_{0}\\v_0=\sqrt{\frac{2gh}{(1-\frac{h^{2}}{R^{2}})^{-2}-1}}$

Using the fact that $h\ll{R}$ , we can apply Taylor's approximation $(1+x)^{n}\approx(1+nx)$ , finally we have:

$v_{0}=R\sqrt{\frac{g}{h}}=14 m/s$