Kinematics

Galileo, Newton, and a number of other renowned natural philosophers contributed to the development of the kinematics formulas. Newton, of course, was the primary contributor to our modern understanding of kinematics. v is velocity, a is acceleration, t is time, and x is displacement. We must always be careful to make sure our units match when doing calculations and that our signs are correct (for instance, gravity is often a negative acceleration with respect to initial vertical velocity).

$v_{final} = v_{initial} + at$

$x = \left(\frac{v_{final} + v_{initial}}{2}\right)t$

$x = v_{initial}t + \frac{1}{2}at^2$

$v_{final}^2 = v_{initial}^2 + 2ax$

Rotational motion can be analyzed in a way that is very similar to our standard kinematics. We have some nearly analogous equations. $\omega$ is the rotational velocity, $\alpha$ is the rotational acceleration, t is time, and $\theta$ is the angular displacement.

Recall that $\omega = \frac{v}{r}$ and $\alpha = \frac{v^2}{r}$.

$\omega_{final} = \omega_{initial} + \alpha t$

$\theta = \left(\frac{\omega_{final} + \omega_{initial}}{2}\right)t$

$\theta = \omega_{initial}t + \frac{1}{2}\alpha t^2$

$\omega_{final}^2 = \omega_{initial}^2 + 2\alpha\theta$

f is the force due to gravity. $m_1$ and $m_2$ are the masses of the two objects in kg. r is the distance between the two objects in meters. G is the constant of universal gravitation or $G \approx 6.674 \cdot 10^{-11} \text{ }\frac{m^3}{kg\cdot s^2}$.

$f = G\frac{m_1m_2}{r^2}$

We have two equations force that are closely associated to, but distinct from our kinematics equations.

$F = ma$

$F = \frac{mv^2}{r} = mr\omega^2$

The first thing we must do is discard irrelevant information. If we are using the kinematics equations they will be a good approximation of the rocket flight path, but we do not need the mass of the rocket or its fuel. We need only the acceleration and the time spent accelerating. We will need to divide the rockets flight into three stages though: stage 1 will be powered acceleration, stage 2 will be flight given an initial velocity with no acceleration, and stage 3 will be free fall.

Stage 1:
We know we spend 30 seconds during stage 1. Let us calculate our height obtained and final velocity during this period. Here, we have use
$x = v_{initial}t + \frac{1}{2}at^2$
$x = \frac{1}{2}(30)30^2 = 13500 \text{ m}$ attained during stage 1

$v_{final} = v_{initial} + at$
$v_{final} = 30(30) = 900 \text{ } \frac{m}{s}$ final velocity attained during stage 1 which will become our initial velocity during stage 2.

Stage 2:
Our initial velocity during stage 2 is our final velocity from stage 1. Our only accelerating force here is gravity which will necessarily have the opposite sign as our initial velocity. We know that stage 2 will end when the final velocity is zero and our rocket begins to fall.
$v_{final} = v_{initial} + at$
$0 = 900 – 9.8t \rightarrow t = \frac{900}{9.8} = 91.84 \text{ s}$ time spent in stage 2 of rocket ascent.

$x = v_{initial}t + \frac{1}{2}at^2$
$x = 900(91.84) + \frac{1}{2}(-9.8)(91.84)^2 =41327 \text{ m}$ height achieved during stage 2 of rocket ascent.

We can now answer the first question. Our rocket obtains a maximum height of $13500 + 41327 \approx 54800 meters$.

Stage 3:
Our rocket is now in free fall and will continue to accelerate until it hits the ground. We will calculate a final velocity here for the sake of the third part of the question though we could approximately graph it without such a thing.
$x = v_{initial}t + \frac{1}{2}at^2$
$54800 = \frac{1}{2}(9.8)t^2 \rightarrow t = \sqrt{\frac{2\cdot 54800}{9.8}} = 105.75 \text{ s}$

$v_{final} = v_{initial} + at$
$v_{final} = -9.8(105.75) = -1036 \text{ } \frac{m}{s}$

We can say the rocket spent a total time in the air of $t_{total} = t_{stage 1} + t_{stage 2} + t_{stage 3} = 30 + 91.84 + 105.75 = 227.59 s$

I’d like to point some things out about the graph that are interesting and useful. We could have solved the entire problem by first making this relatively elementary graph and finding the area of the blue triangle. That area will equal the total displacement of our rocket upwards. Next, we continue to line segment until we find a point such that area of the pink triangle equals the area of the blue triangle. There our negative and positive displacement are equal meaning the rocket has returned to the ground.

There are some approximations in this problem that will break down under close scrutiny. The most obvious one is air resistance, but there is also an issue with other parameters. For instance, gravity weakens as you go far from the Earth! Even at a height of 54 km, this weakening is small, but it is definitely measurable.

We know the first pebble has $v_{initial} = 2.0 \text{ }\frac{m}{s}$, $a = 2.0 \text{ }\frac{m}{s^2}$, and $x = 50 \text{ m}$. We need to determine the time of flight for the first pebble from this. Note that gravity and the initial velocity have the same sign in this problem because they are in the same direction.

We apply the following kinematics equation:
$x = v_{initial}t + \frac{1}{2}at^2$
$50 = 2t + \frac{1}{2}(9.8)t^2$
We have a quadratic equation and we apply the quadratic formula to obtain:
$4.9t^2 + 2t – 50 = 0 \rightarrow t = 3.0 \text{ s, } -3.4 \text{ s}$
3.0 s is the flight of the first pebble and we determine that since there was only one splash, we may assume both pebbles hit the water at the same time. Since the second pebble was thrown 1.00 seconds later, it was in the air for 3.0 – 1.0 = 2.0 s.

We know the flight time of the second pebble, the height from which it was thrown, and its acceleration. We again utilize:
$x = v_{initial}t + \frac{1}{2}at^2$
$50 = v_{initial}(2) + \frac{1}{2}(9.8)(2)^2 \rightarrow v_{initial} =15.2 \text{ } \frac{m}{s}$
Now, we have the initial velocity and the time of flight so we can calculate the final velocity
$v_{final} = v_{initial} + at$
$v_{final} = 15.2 + 9.8(2) = 34.8 \approx 35 \text{ }\frac{m}{s}$

The final velocity of the second pebble is $35 \text{ }\frac{m}{s}$ downwards and its flight time is 2.0 s.

This graph is also profoundly useful were you to be provided with it prior to working the problem. The slope at any point (the derivative of the displacement) yields the velocity. We can see the graph is quite steep at t = 2.0 s corresponding to the final velocity of 35 $\frac{m}{s}$. The second derivative of this graph yields the acceleration which would be the constant, g.

$F = ma_c \rightarrow$ Recall that $a_c = \frac{v^2}{r}$
$F = \frac{mv^2}{r} \rightarrow$
$\frac{Fr}{m} = v^2 \rightarrow$
$\sqrt{\frac{Fr}{m}} = v \rightarrow \sqrt{\frac{57(1.4)}{.45}} = 13 \text{ }\frac{m}{s}$ noting that we need 2 significant figures!

The first issue we run into here is that we need to convert all the units to proper units for use in our rotational motion equations. We also need to understand what the net force diagram looks like.

Since the coin is not moving, we start with:

$F_{friction} = F_a$
$\mu_s mg = ma_c$
$\mu_s mg = m\frac{v^2}{r}$
$\mu_s \cancel{m}g = \cancel{m}\frac{v^2}{r}$
$\mu_s = \frac{v^2}{rg} \rightarrow \frac{1.0^2}{(.24)(9.8)} = .43$

$F_{friction} = F_a$
$\mu_s mg = ma_c$
$\mu_s mg = m\frac{v^2}{r}$
$\mu_s \cancel{m}g = \cancel{m}\frac{v^2}{r}$
$\sqrt{\mu_srg} = v \rightarrow \sqrt{.33*.85*9.8} = 1.7 \text{ }\frac{m}{s}$

$F_g = F_a$
$mg = ma_c$
$\cancel{m}g = \cancel{m}a_c$
$g = a_c$
$g = \frac{v^2}{r}$
$\sqrt{gr} = v \rightarrow \sqrt{9.8*1.3} = 3.5 \text{ }\frac{m}{s}$

$F_{\text{gravity of new planet}} = F_{Earth}$

$\frac{Gm_{\text{new planet}}m_{probe}}{r_{\text{new planet}}^2} =\frac{Gm_{Earth}m_{probe}}{r_{Earth}^2}$

$\frac{G2m_{Earth}m_{probe}}{r_{\text{new planet}}^2} =\frac{Gm_{Earth}m_{probe}}{r_{Earth}^2}$

$\frac{2\cancel{Gm_{Earth}m_{probe}}}{r_{\text{new planet}}^2} =\frac{\cancel{Gm_{Earth}m_{probe}}}{r_{Earth}^2}$

$\frac{2}{r_{\text{new planet}}^2} =\frac{1}{r_{Earth}^2}$

$r_{\text{new planet}}^2=2r_{Earth}^2$

$r_{\text{new planet}}=\sqrt{2}r_{Earth}$

The problem begs us to look up a number of numerical constants which are readily available via google. We also need to recall that to calculate the orbital altitude, we must consider this altitude from the center of the Earth- therefore we need to include the radius of the earth + orbital height.

$F_a = F_g$

$m_{satellite}a_c = \frac{Gm_{satellite}m_{Earth}}{(r_{earth}+1000000)^2}$

$m_{satellite}\frac{v^2}{(r_{earth}+1000000)} = \frac{Gm_{satellite}m_{Earth}}{(r_{earth}+1000000)^2}$

$\cancel{m_{satellite}}\frac{v^2}{(r_{earth}+1000000)} = \frac{G\cancel{m_{satellite}}m_{Earth}}{(r_{earth}+1000000)^2}$

$\frac{v^2}{\cancel{(r_{earth}+1000000)}} = \frac{Gm_{Earth}}{(r_{earth}+1000000)^\cancel{2}}$

$v^2 = \frac{Gm_{Earth}}{r_{earth}+1000000}$

$v = \sqrt{\frac{Gm_{Earth}}{r_{earth}+1000000}} = \sqrt{\frac{6.674 \cdot 10^{-11} 5.97219 \cdot 10^{24}}{6378100+1000000}} = 7350.0 \text{ }\frac{m}{s}$

Here, we have to consider the net forces thoughtfully. We will have two orthogonal sums here:

$\Sigma F_x = ma$ and $\Sigma F_y = 0$.

The second equation is zero precisely because our assumption is that we are going the fastest possible speed that does NOT allow for upwards slipping.

$-N_x – f_x = -ma$ and $N_y – W – f_y = 0$

$Nsin(\theta)+ \mu Ncos(\theta) = m \frac{v^2}{r}$ and $Ncos(\theta) – mg – \mu Nsin(\theta) = 0$

Rewriting the second equation, we obtain:

$N(cos(\theta) – \mu sin(\theta)) = mg$

$N= \frac{mg}{cos(\theta) – \mu sin(\theta)}$

And rewriting the first equation to solve for v, we get:
$v = \sqrt{\frac{N \cdot r(sin(\theta) + \mu cos(\theta))}{m}}$

Substituting for N, we have
$v = \sqrt{\frac{\left(\frac{mg}{sin(\theta) + \mu cos(\theta)}\right) \cdot r(cos(\theta) – \mu sin(\theta))}{m}}$

$v = \sqrt{\frac{gr(sin(\theta) + \mu cos(\theta))}{cos(\theta) – \mu sin(\theta)}}$

$v = \sqrt{\frac{9.8(26)(sin(16) + .58 cos(16))}{cos(16) – .58 sin(16)}} \approx 16.2759 \text{ }\frac{m}{s}$