In the video above, I calculate the friction inside of the barrel.

In the following paragraphs, I explain several different methods
that can be used to calculate this friction. When I shoot the ball,
the ball hits the sides of the barrel.

I know that friction is a force, and that there are two types of friction: static friction and kinetic friction. Static friction prevents an object from moving; however, kinetic friction is present when an object is moving.

The static friction \(f_s\) is defined as a constant multiplier \(\mu_s\)
times the normal force N; in another words: \(f_s = \mu_s N \).

Similarly, the kinetic friction \(f_k\) is defined as a constant
multiplier \(\mu_k\) times the normal force N; in another words:
\(f_k = \mu_k N \).

Note that friction always act against an applied
force, which is why its sign is negative. Therefore, I must subtract
the friction force to get our net force. Below equations is a summarized
version of what I said.

Static friction can only be equal to or less than the force applied to the
object; however, the kinetic friction is always less than the applied force.

In the below diagram, as the applied force (horizontal-axis) to the
object increases, the static friction (vertical-axis) increases till a
breaking point where the object starts moving; at this moment, I are dealing
with kinetic friction. Unlike static friction, increasing the applied force does not affect the force of kinetic friction.

In the above image, it shows an object on a slope, but it is
not moving because there is static friction. The diagonal downward force
applied by gravity is \(f = m g \ sin \ \theta\); the friction force acting against
it in the opposite direction is \(f = \mu N\).

To do some calculations, I will need to figure out alpha
\(alpha\),
which is same as beta \(\beta\). To verify this, I start by
with the following expressions for a right triangle.

Now that I know beta, I can solve for alpha by substitute beta into the equation.

$$ \alpha = 90 - \beta \\ \alpha = 90 - (90 - \beta) \\ \alpha = 90 - 90 + \beta \\ \alpha = \theta $$Now that I have determined that \( \alpha = \theta \), I will calculate the force \(F\) of the object going down the slope and the normal force \(N\) with mass \(m\) and gravity \(g\).

$$ sin \theta = \frac{F}{mg} \\ F = m g \ sin \theta \\ cos \theta = \frac{N}{mg} \\ N = m g \ cos \theta $$Now that I have the force and the normal force, I can calculate our constant friction multiplier, \(\mu_s\).

$$ \sum F = f - f_s = 0 \\ m g \ sin \theta - \mu_s N = 0 \\ m g \ sin \theta = \mu_s m g \ cos \theta \\ \mu_s = \frac{m g \ sin \theta}{m g \ cos \theta} \\ \mu_s = tan \theta $$I now know the static multiplier \(\mu_s\) for a given angle; but I want to find the kinetic friction for the object. Below are several ways to calculate the kinetic multiplier \(\mu_k\).

I came with three ways to calculate the friction inside the barrel.

- Balls sliding down the barrel
- Using two different barrels
- Using different angles on the same barrel

The sections that follow describe above points in detail.

The first way to calculate the kinetic friction uses a method that requires
a ball sliding down a slope. Since a ball rolls down a slope, I can attach two
balls with tape and measure how long it takes them to slide down the barrel.

I can now calculate the acceleration of the object:

Now that I know our acceleration, I can calculate the kinetic friction multiplier \(\mu_k\).

$$ \sum F = F - f_k \\ m a = m g \ sin \theta - \mu_k (N) \\ \frac{\cancel{m}a}{\cancel{m}} = \frac{ \cancel{m} g \sin{\theta} - \mu_k (\cancel{m} g \ cos \theta) }{\cancel{m}} \\ a = g \ sin \theta - \mu_k g \ cos \theta \\ a + \mu_k g \ cos \theta = g \ sin \theta \\ \mu_k g \ cos \theta = g \ sin \theta - a \\ \frac{\mu_k \cancel{g \ cos \theta}}{\cancel{g \ cos \theta}} = \frac{g \ sin \theta -a}{g \ cos \theta} \\ \mu_k = \frac{\cancel{g} sin \theta}{\cancel{g} \ cos \theta} - \frac{a}{g \ cos \theta} \\ \mu_k = tan \theta - \frac{a}{g \ cos \theta} $$Assuming I already experimented to determine the velocity with friction by shooting horizontally I have the equation \(v=\sqrt{\frac{g \ x^2}{2y}}\). Now, I want to determine the velocity without friction (e.g. at the beginning of the barrel) so that I can determine final velocity (including friction) for a given angle. Given that:

$$ \begin{align*} f_k &= \mu_k \ N \\ &= \mu_k \ m \ g \ cos \theta \\ &= \mu_k \ m \ g \ cos 0^o \\ m a &= \mu_k\ m g \\ a &= \mu_k \ g \end{align*} $$Therefore the velocity without friction at the beginning of the barrel is:

$$ v_{nf}^2 = v_o^2 + 2 l a \\ v_{nf}^2 = \frac{g \ x^2}{2y} + 2 l (\mu_k \ g) $$Now I can add the friction deceleration for a given angle.

$$ \begin{align*} v_f^2 &= (v_{nf}^2) - 2 l (a) \\ v_f^2 &= (\frac{g \ x^2}{2y} + 2 l \ \mu_k \ g) - 2 l (\mu_k \ m \ g \ cos \theta \\ v_f^2 &= \frac{g \ x^2}{2y} + 2l \ \mu_k \ g (1 - cos \ \theta) \end{align*} $$
I will calculate the kinetic friction using two barrels, which may
be more accurate because the ball bounces on the barrels instead of
sliding down the barrels.

A picture of what I are working with is shown below:

First thing I want to do is calculate how much more deceleration there is in the longer barrel (barrel 1 in above image)than there is in the shorter barrel (barrel 2 in above image). Note that the end tip of the barrels are in the same loccation; but, they have different starting points. The velocity \(v_1\) is for the first (shorter) barrel, and \(v_2\) is the velocity for the second (longer) barrel. First, I must calculate the velocity for each barrel. Next, I can determine the deceloration between the two.

$$ v_1^2 = v_2^2 + 2 a_1 l \\ v_1^2 - v_2^2 = 2 a_1 l \\ a_1 = \frac{v_1^2 - v_2^2}{2 l} $$Since I have enough information to calculate our \(\mu_k\), I might as well calculate it right now.

$$ F = m a_1 = \mu_k (N) \\ F = m a_1 = \mu_k (m g \ cos \theta) \\ \frac{\cancel{m} a_1 }{\cancel{m} g \ cos \theta} = \frac{\mu_k \cancel{m g \ cos \theta}} {\cancel{m g \ cos \theta}} \\ \mu_k = \frac{a_1}{g \ cos \theta} = \frac{a_1}{g \ cos \ 90^o} = \frac{a_1}{g} \tag{1} $$Now I will solve for velocity with no friction (e.g. \(v_{nf}\)) at the beginning of the barrel, so that I can determine the final velocity for a given angle.

$$ V_{nf}^2 = v_1^2 + 2 a_1 l \\ V_{nf} = \sqrt{v_1^2 + 2 a_1 l} $$
Now I will calculate the final velocity; considering the angle of the barrel and the friction applied.

Note that equation (1) for \(\mu_k\ = \frac{a_1}{g \ cos \theta}\); therefore, friction decelration at a given angle is \(a_2 = \mu_k g \ cos \theta\).

Using the same formula I used for calculating the cannon pitch angle, I will calculate two different formulas, and use one to fill in the blanks for the other.

First of all, I want figure out a formula that I could use to calculate the cannon velocity. Even though I have calculated the cannon velocity in a different link, it did not include kinetic friction. Below, I will show a way to calculate with friction. I start with a projectile motion formula.

$$ y = v_y t + \frac{1}{2} a t^2 \\ x = v_x t + \frac{1}{2} a t^2 $$I can substitute the values for the velocities. Unlike Calculating the Cannon Pitch Angle web page, I will not try and calculate the angle. Instead, I will calculate the \(\mu_k\) that I will need.

$$ y = v sin \theta t - \frac{1}{2} g t^2 \\ x = v cos \theta t \tag{3} $$I will solve for time using the above equation:

$$ x = v \ cos \theta \ t \\ t = \frac{x}{v \ cos \theta} $$Now that I have the value for time, I will substitute it into the other equation from (3).

$$ y = v sin \theta (\frac{x}{v cos \theta}) - \frac{1}{2} g (\frac{x}{v cos \theta})^2 $$Through experimentation, I can shoot the projectile and measure the distance from the tip of the cannon barrel to the landing area, which is the distance \(x\). I also know the height of the target platform from the tip of the cannon barrel, to determine the vertical distance \(y\). With these two measurements \(x,y\), I can calculate the projectile velocity; thus, I need to solve for it as following:

$$ y = v sin \theta (\frac{x}{v cos \theta})- \frac{1}{2} g (\frac{x}{v cos \theta})^2 \\ y = \frac{x \cancel{v} sin \theta}{\cancel{v} cos \theta} - \frac{2 x^2}{g v^2 cos^2 \theta} \\ y = x tan \theta - \frac{2 x^2}{g v^2 cos^2 \theta} \\ y - x tan \theta = - \frac{2 x^2}{g v^2 cos^2 \theta} \\ v^2 (y - x tan \theta) = \frac{2 x^2}{g \cancel{v^2} cos^2 \theta}\cancel{v^2} \\ v^2 = \frac{2 x^2}{(y - x tan \theta)g cos^2 \theta} \\ v = \sqrt{ \frac{2 x^2}{(y - x tan \theta)g cos^2 \theta}} $$Now that I have my velocity, I will calculate my deceleration with friction \(\mu_k\).

$$ m a = \mu_k N \\ m a = \mu_k m g \ cos \theta \\ a = \mu_k g cos \theta \\ \tag{4} $$Because I am shooting at two different angles, I will have two different velocities. Velocities differs due to friction forces caused by the barrel angle, which produces different normal forces to the projectile. The velocity at the bottom of the barrel, regardless of the angle, will be the same since no friction is applied to it, we will call this \(V_{nf}\). We solve for our friction by equating the \(V_{nf}\) for the velocity for each angle.

$$ v^2 = v_o^2 + 2 l (a) \\ V_{nf}^2 = v_2^2 + 2 l (\mu_k g cos \boldsymbol \theta_2) \\ V_{nf}^2 = v_1^2 + 2 l (\mu_k g cos \boldsymbol \theta_1) \\ v_1^2 + (2 l \mu_k g cos \theta_2) = v_2^2 + 2 l (\mu_k g cos \theta_1) \\ v_1^2 - v_2^2 = 2 l \mu_k g cos \theta_2 - 2 l \mu_k g cos \theta_1 \\ \frac{v_1^2 - v_2^2}{2 l g} = \mu_k cos \theta_2 - \mu_k cos \theta_1 \\ \frac{v_1^2 - v_2^2}{2 l g} = \mu_k (cos \theta_2 - cos \theta_1) \\ \mu_k = \frac{\frac{v_1^2 - v_2^2}{2 l g}}{(cos \theta_2 - cos \theta_1)} \\ \mu_k = \frac{v_1^2 - v_2^2}{2 l g (cos \theta_2 - cos \theta_1)} \tag{5} $$Similar to equation (2), I can now calculate the final velocity; considering the angle of the barrel and the friction applied. We use non-friction velocity \(v_{nf}\) from equation (5) and the friction deceleration from equation (4); \(\theta_2\) was from experimentation where \(\theta\) is the target angle. In the following expression:

$$ v_f^2 = V_{nf}^2 - 2 l (a) \\ v_f^2 = \Big(v_2^2 + 2 l (\mu_k g cos \boldsymbol \theta_2)\Big) - 2 \ l \ (\mu_k \ g \ cos \theta) \\ \tag{6} $$
That is how I calculate the friction of the barrel.

Just in case you still don't have a very good understanding of friction yet,
click here.