Now pick it up

First, we will assume that the collision wih the floor happens in a very short amount of time. We will assume that the pencil moves towards the floor at a constant velocity $v_0$ prior to the collision. We will assume that during the collision, a constant force of intensity $F$ acted upwards on one end of the pencil during the time $\Delta t$ . After the collision, let $\omega$ denote the angular velocity of the pencil and let $v$ denote the now upward velocity of the pencil. The mass of the pencil is $m$ , its moment of inertia with respect to the axis of rotation is $I_0$ and its half-length is $r$ . We will use $M$ to denote the torque acting on the pencil during the collision, and finally $\alpha$ to denote the initial angle of the pencil with respect to the horizontal.

We know that the total mechanical energy before and after the collision stays unchanged. Using the well-known formulas for kinetic and rotational energy we arrive at our first equation:

$\frac{1}{2}mv_0^2 = \frac{1}{2}I_o\omega^2 + \frac{1}{2}mv^2$

Now we have Newton's second law for rotation:

$I_o\omega = M\Delta t$

And finally Newton's second law for linear motion:

$F\Delta t = mv + mv_0$

Notice that we wrote Newton's second law in this form because the velocity changed from downward $v$ to upward $v_0$ . Also, we have:

$M = F r \cos \alpha$

Now we just solve this system of equations for the variable $\omega$ step by step:

First, we eliminate $M$ by substituting $M = F r \cos \alpha$ into our second equation:

$I_o\omega = F \Delta t r \cos \alpha$

We can also eliminate $F \Delta t$ by substituting the third equation into the second:

$I_o\omega = m(v + v_0) r \cos \alpha$

Now we eliminate $I_0$ using the well-known formula for the moment of inertia $I_0 = \frac{1}{3} m r^2$ , and then simplify the equations a little bit. We have:

$v_0^2 = \frac{1}{3} r^2 \omega^2 + v^2$

and

$\frac{1}{3} r \omega = (v + v_0) \cos \alpha$

We solve the second equation for $v$ :

$\frac{r \omega}{3 \cos \alpha} - v_0 = v$

And substitute right into the first equation:

$v_0^2 = \frac{1}{3} r^2 \omega^2 + (\frac{r \omega}{3 \cos \alpha} - v_0)^2$

Now we just solve this for $\omega$ :

$(\frac{r \omega}{3 \cos \alpha} - v_0)^2 = \frac{r^2 \omega^2}{9 \cos^2 \alpha} + v_0^2 - \frac{2 r \omega v_0}{3 \cos \alpha}$

$v_0^2 = \frac{1}{3} r^2 \omega^2 + \frac{r^2 \omega^2}{9 \cos^2 \alpha} + v_0^2 - \frac{2 r \omega v_0}{3 \cos \alpha}$

$0 = \frac{1}{3} r^2 \omega^2 + \frac{r^2 \omega^2}{9 \cos^2 \alpha} - \frac{2 r \omega v_0}{3 \cos \alpha}$

$\frac{2 r \omega v_0}{3 \cos \alpha} = \frac{1}{3} r^2 \omega^2 + \frac{r^2 \omega^2}{9 \cos^2 \alpha}$

We will now eliminate one $\omega$ from the equation. We can do this since we know it is not equal to zero.

$\frac{2 r v_0}{3 \cos \alpha} = \frac{1}{3} r^2 \omega + \frac{r^2 \omega}{9 \cos^2 \alpha}$

This is now a ordinary linear equation in $\omega$ . First, I will eliminate fractions by multiplying both sides with $9 \cos^2 \alpha$ .

$6 r v_0 \cos \alpha = 3 r^2 \omega \cos^2 \alpha + r^2 \omega$

$6 r v_0 \cos \alpha = \omega(3 r^2 \cos^2 \alpha + r^2)$

$\omega = \frac{6 r v_0 \cos \alpha}{3 r^2 \cos^2 \alpha + r^2}$

$\omega = \frac{1}{r} \frac{6 v_0 \cos \alpha}{3 \cos^2 \alpha + 1}$

Since $r$ , $v_0$ and $6$ are positive constants, we can eliminate them from the expression on the right ant then search for the maximum value of a simpler expression:

$f = \frac{\cos \alpha}{3 \cos^2 \alpha + 1}$

Substituting $x = \cos \alpha$ and restricting $-1 \leq x \leq 1$ , now we can search for the maximum value of an even simpler expression:

$f = \frac{x}{3 x^2 + 1}$

The first derivative of this expression should be equal to zero in a local extremum:

$f' = \frac{1(3 x^2 + 1) - 6x^2}{(3 x^2 + 1)^2} = 0$

$f' = \frac{1 - 3 x^2}{(3 x^2 + 1)^2} = 0$

Now we simply solve:

$1 - 3 x^2 = 0$

to get

$x = \frac{1}{\sqrt 3}$

Notice that we only need a positive solution for $x$ . This truly is a local maximum, it can be verified by calculating the sign of the second derivative of the original expression. To get the original angle we can use the inverse cosine function.

$\alpha = \arccos \frac{1}{\sqrt 3} \approx 54.74^\circ$

I think the assumption of "perfectly elastic" is redundant. In my solution, I do not use the assumption of "perfectly elastic", although it can slightly simplify the workings. Let the length of the pencil be 2 $d$ and the mass be $m$ . Let the angle the pencil makes with the horizontal be $\theta$ . Assume the time of contact of the pencil and the ground is $t$ . Since the collision usually takes place in a very short time, we assume $t$ is small enough such that $\theta$ does not change throughout the collision. Assume the center of mass of the pencil is moving vertically downward with speed $v_0$ just before the collision and is moving downward with speed $v$ and angular speed $\omega$ (around the axis of its center of mass) just after the collision. When the pencil is rotating, the velocity of the end of the pencil in the rest frame is determined by (velocity of the end in the frame of center of mass) $+$ (velocity of the center of mass). Hence, the vertical speed (assume up is positive) of the end touching the ground can be determined by (vertical speed of the end in the center of mass frame) $-$ (downward vertical speed of the center of mass). The vertical speed of the end just after the collision $= d \omega \cos \theta$ , because the speed of the end is given by the pencil's angular speed $\omega$ times the distance between the end and its center of mass $d$ and the speed is making an angle of $\theta$ with the vertical. Just after the collision, the center of mass is moving downward with speed $v$ , by assumption. At the moment of the end leaving the ground, the vertical speed of the end is 0, so $d \omega \cos \theta - v = 0$ Assume the force (only in the vertical direction since there is no friction) exerting on the pencil by the ground throughout the collision is F. Note that F is changing during the collision. Assume that F is far greater than the pencil's weight, so the sole force acting on the pencil during the collision is F, upward. During the time $t$ , vertical linear impulse, $\int F\, \mathrm{d}x = m (v_0 - v)$ , where $m (v_0 - v)$ is the change in vertical linear momentum. The total amount of torque the F makes about the center of mass is $\int F d \cos \theta \, \mathrm{d}x = I \omega$ , where $I$ is the moment of inertia of the pencil about its center of mass ( $I = \frac {1}{12} m (2d)^2 = \frac {1}{3} m d^2$ ) and $I \omega$ is the change in angular momentum. Since $d \cos \theta$ is constant, we can take the two terms out of the integral and divide by them to get $\int F \, \mathrm{d}x = \frac{I \omega}{d \cos \theta} = \frac{m d \omega}{3 \cos \theta}$ From $d \omega \cos \theta - v = 0$ and $\int F\, \mathrm{d}x = m (v_0 - v)$ , we get $\int F\, \mathrm{d}x = m (v_0 - d \omega \cos \theta)$ . Since $\int F \, \mathrm{d}x = \frac {m d \omega}{3 \cos \theta}$ , $m (v_0 - d \omega \cos \theta)=\frac {m d \omega}{3 \cos \theta}$ Cancel $m$ : $v_0 - d \omega \cos \theta=\frac {d \omega}{3 \cos \theta}$ . Rearrange and we get the expression for $\omega$ : $\omega = \frac{v_0 \cos \theta}{d (\frac {1}{3} + \cos^2 \theta)}$

Suppose that the rod has mass $m$ , length $2L$ , and that it is falling with speed $v$ when it hits the ground, making an angle of $\theta$ with the horizontal. Suppose that, after the collision, the centre of mass of the rod has downward speed $V$ , and that the rod has angular velocity $\omega$ .

During the collision, the only impulses act at the point of contact of the rod with the floor. Thus angular momentum (about that point of contact) is preserved, which implies that $\tfrac13mL^2\omega + mVL\cos\theta \; = \; mvL\cos\theta$ Since the collision is perfectly elastic, kinetic energy is conserved during the collision. Thus $\tfrac16mL^2\omega^2 + \tfrac12mV^2 \; = \; \tfrac12mv^2$ Solving these equations yields $\begin{array}{rcl} \displaystyle 3(v^2 - V^2) & = & \displaystyle L^2\omega^2 \; = \; 9(v-V)^2\cos^2\theta \\ \displaystyle v+V & = & 3(v-V)\cos^2\theta \\ \displaystyle V & = & \frac{3\cos^2\theta - 1}{3\cos^2\theta + 1}v \end{array}$ and hence $\omega \; = \; \frac{6v}{L} \frac{\cos\theta}{3\cos^2\theta +1}$ which is maximized when $\cos\theta = \tfrac{1}{\sqrt{3}}$ , so when $\theta = \boxed{54.73561032^\circ}$ .

Let us assume that we look at the system from C.M. view. assume that the length of the pencil is 2L, so the angular Mom. inertia is m(2L)^2/12=M(L^2)/3.

Assume that the pencil is impacting at vertical speed of V1 and bounces at speed V2. And the resulting angular speed of W .

So the pencil which hits the earth will experience upon impact velocity change of V2-V1 which from the point of the C.M will give an momentum change=Impact Force of about M(V2-V1)xLxCos(A) where A is the angle of the pencil with the horizon. So we can write as the total Ang. Momntum has to be 0 as at the start. -IxW=Mx(V2-V1)xLxCos(A) Eq.1

From the point of energy conservation.

Impact energy=1/2M(V1)^2=1/2M(V2)^2+1/2IxW^2=Recoil energy+rotation energy

we can write(EQ.1) V1-V2=Ixw/(MxLxcos(A)) Which can be substituted into the energy Eq. above to give (1/2MxIxW/MLcos(A))x(V1+V2)=1/2xIxW^2

so we have 2 Eq. V1+V2=WxLxcos(A) and And we have also -IxW=Mx(V2-V1)xLxCos(A) the Final result is (too much to elaborate) W=2V1MLcos(A)/(M(lxcos(A)^2+I) To find the maximum we take the derivative and compare to 0. Sin(A)/(cos(A)^2+1/3)[-1+2cos^2(A)x/(Cos(A)^2+1/3)] EQ.2

One solution is A=0 obviously a pure bounce without rotation, as no side of the pencil has preference on the other side.

So this can't be the maximum we look for. The other is from the condition inside the square brackets of Eq.2....cos(A)^2=1/3 or cos(A)=SQRT(1/3) so Arccos(SQRT1/3)=54.7356 degrees.

BTW, where can I find the solutions for questions from previous weeks. Thanks ahead

Let initial velocity=v,angle to horizontal=θ Let final velocity=v',angular velocity=ω Energy conserved.For maximum ω,v'should be zero. So, ½ mv^2= ½ ml^2/12 ω^2 --------1 so,impulse=mv angular impulse=mvlcosθ/2 = ml^2/12 ω---------2 solving we get θ

Firstly , we realize that collision is elastic that means we can conserve energy of the pencil during the process. Secondly , We use the fact that elastic collision implies that the velocity of point of contact along the normal of collision(that is here normal of the surface) gets negative but with same magnitude or reverses its direction. Let us assume the final configuration of the rod. Let the rod rotate with some angular velocity and it has some velocity of centre of mass v after the collision. Let u be the initial velocity of centre of mass.

$\frac{1}{2}mu^2=\frac{1}{2}I{\omega}^2+\frac{1}{2}mv^2$ $\frac{l\omega}{2}\cos\theta+v=u$ Now , we have two unknowns that are v and angular velocity. Solving for angular velocity $\omega=\frac{12u\cos\theta}{l+3{l\cos^2\theta}}$ Differentiate angular velocity with respect to theta and taking the derivative as 0. We get , $cos2\theta=-\frac{1}{3}$ Solving for theta , $\theta=\cos^{-1}(\frac{1}{\sqrt{3}})$