4.2: Dynamics

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Concepts and Principles

We are now prepared to investigate forces that are free to vary in magnitude and direction. However, you may recall that at no time in the explanation of Newton’s laws was the conversation restricted to either constant acceleration or constant force. In fact, Newton’s laws are completely valid regardless of the nature of the forces acting on an object, or its acceleration.⁴ Thus, the relation

\[\Sigma F=m a \nonumber\]

remains valid, and is still the central relationship in the study of mechanics.

Note

⁴ Actually, Newton’s Second Law is known to be inadequate to explain phenomenon occurring at extremely high speeds or at extremely small distances. The speeds must be comparable to the speed of light or the distances must be comparable to the size of the atom for these effects to be noticeable.

Analysis Tools

Time-Dependent Forces

A 2.0 kg toy rocket is fitted with an engine that provides a thrust roughly modeled by the function F(t) = (60 N/s) t - (15 N/s²) t², for 0 < t < 4.0 s, and zero thereafter. The rocket is launched directly upward.

Of course, the first step to analyzing any situation involving forces is to construct a free-body diagram. Below is a free-body diagram for the rocket during the time interval 0 < t < 4.0 s.

Given that you are not presented with information pertaining to the frictional force acting on the rocket, you must analyze the rocket’s trajectory in a hypothetical, friction-free environment.

From Newton’s second law:

\[\begin{aligned}
&\Sigma F=m a \\
&F_{\text {exhaust }}-F_{\text {gravity }}=m a \\
&60 t-15 t^{2}-2(9.8)=2 a \\
&-7.5 t^{2}+30 t-9.8=a
\end{aligned} \nonumber\]

Thus, the acceleration of the rocket is not constant, and the velocity and position of the rocket must be determined by integrating the acceleration.

Before we do this, however, note that the acceleration of the rocket is initially directed downward. (At t = 0s, a = -9.8 m/s².) This is because the supporting force exerted on the rocket by the launch platform has been ignored. This force would hold the rocket in place until the force of the exhaust gases on the rocket is equal to, and then exceeds, the force of gravity on the rocket. In reality, the acceleration of the rocket is zero until F_exhaust = F_gravity. This occurs when:

\[\begin{aligned}
&F_{\text {exhaust }}=F_{\text {gravity }} \\
&60 t-15 t^{2}=2(9.8) \\
&-15 t^{2}+60 t-19.6=0 \\
&t=0.36 \mathrm{~s}
\end{aligned} \nonumber\]

The correct motion information is tabulated below.

Event 1: The engine is ignited	Event 2: Rocket leaves the pad	Event 3: The thrust ends	Event 4: The rocket reaches its apex
t₁ = 0 s	t₂ = 0.36 s	t₃ = 4.0 s	t₄ =
r₁ = 0 m	r₂ = 0 m	r₃ =	r₄ =
v₁ = 0 m/s	v₂ = 0 m/s	v₃ =	v₄= 0 m/s
a₁ = 0 m/s²	a₂ = 0 m/s²	a3 =	a₄= -9.8 m/s²

Between event 2 and 3, the rocket’s acceleration is given by the function above:

\(a(t)=-7.5 t^{2}+30 t-9.8\)

We can integrate the acceleration to determine the velocity,

\(\begin{aligned}
&v(t)=\int a(t) d t \\
&v(t)=\int\left(-7.5 t^{2}+30 t-9.8\right) d t \\
&v(t)=-2.5 t^{3}+15 t^{2}-9.8 t+C
\end{aligned}\)

Since we know v = 0 m/s when t = 0.36 s, we can determine the integration constant:

\(\begin{aligned}
&v(0.36)=-2.5(0.36)^{3}+15(0.36)^{2}-9.8(0.36)+C=0 \\
&C=1.70 \mathrm{~m} / \mathrm{s}
\end{aligned}\)

Therefore,

\(v(t)=-2.5 t^{3}+15 t^{2}-9.8 t+1.7\)

and when the thrust ends, at t₃ = 4.0 s, v₃ = 42.5 m/s.

To find the position of the rocket when the thrust ends, integrate the velocity function:

\(\begin{aligned}
&r(t)=\int v(t) d t \\
&r(t)=\int\left(-2.5 t^{3}+15 t^{2}-9.8 t+1.7\right) d t \\
&r(t)=-0.625 t^{4}+5 t^{3}-4.9 t^{2}+1.7 t+D
\end{aligned}\)

Since we know r = 0 m when t = 0.36 s, we can determine the integration constant:

\(\begin{aligned}
&r(0.36)=-0.625(0.36)^{4}+5(0.36)^{3}-4.9(0.36)^{2}+1.7(0.36)+D=0 \\
&D=-0.178
\end{aligned}\)

Therefore,

\(r(t)=-0.625 t^{4}+5 t^{3}-4.9 t^{2}+1.7 t-0.178\)

and when the thrust ends, at t₃ = 4.0 s, r₃ = 88.2 m.

Between event 3 and 4 the acceleration is constant, so we can use our constant-acceleration kinematic equations:

\[ \begin{array}{ll}
\mathrm{v}_{4}=\mathrm{v}_{3}+\mathrm{a}_{34}\left(\mathrm{t}_{4}-\mathrm{t}_{3}\right) & \mathrm{r}_{4}=\mathrm{r}_{3}+\mathrm{v}_{3}\left(\mathrm{t}_{4}-\mathrm{t}_{3}\right)+1 / 2 \ \mathrm{a}_{34}\left(\mathrm{t}_{4}-\mathrm{t}_{3}\right)^{2} \\
0=42.5+(-9.8)\left(\mathrm{t}_{4}-4\right) & \mathrm{r}_{4}=88.2+42.5(8.34-4)+1 / 2 \ (-9.8)(8.34-4)^{2} \\
\mathrm{t}_{4}=8.34 \mathrm{~s} . & \mathrm{r}_{4}=180 \mathrm{~m}
\end{array} \nonumber \]

Thus, the rocket would achieve a maximum height of 180 m in a friction-free environment.

Applying Newton's Second Law to Circular Motion

Investigate the situation below, in which an object travels on a curved path.Note

During a high speed auto chase in San Francisco, a 1745 kg police cruiser traveling at constant speed loses contact with the road as it goes over the crest of a hill with radius of curvature 80 m.

A free-body diagram for the car at the instant it’s at the crest of the hill is sketched below.

It’s doubtful that the shape of the hill that the car travels on is circular. However, at every point on the hill we can approximate the path of the car with a circle of appropriate radius. This “instantaneous circular path” has a radius equal to the given radius of curvature. Any curved path can be approximated as a series of circular paths. Therefore, since the car is traveling along a circular path, it’s best to analyze the situation using polar coordinates. Remember, in polar coordinates the radial direction points away from the center of the circular path.

In general you should apply Newton’s second law independently in the radial (r) and tangential (t) directions, although since the cruiser is moving at constant speed there is no acceleration in the tangential direction and therefore no net force in the tangential direction.

\[\begin{aligned} & \qquad\underline{\mathbf { r-direction }}\\ &\begin{aligned} &\Sigma F=m a \\ &F_{\text {road }}-F_{\text {gravity }}=m a_{r} \\ &F_{\text {road }}-F_{\text {gravity }}=m\left(-R \omega^{2}\right) \end{aligned} \end{aligned} \nonumber\]

Since the cruiser loses contact with the road, F_road = 0 N.

\[\begin{aligned}
&0-1745(9.8)=-1745(80) \omega^{2} \\
&\omega=0.35 \mathrm{rad} / \mathrm{s}
\end{aligned} \nonumber\]

Since v = R \(\omega\),

\[\begin{aligned}
&v=(80 \mathrm{~m})(0.35 \mathrm{rad} / \mathrm{s}) \\
&v=28 \mathrm{~m} / \mathrm{s}
\end{aligned} \nonumber\]

For the car to lose contact with the road, it must be traveling at a speed of 28 m/s or greater when it reaches the crest of the hill. At lower speeds, the car would remain in contact with the road.

Since it is often useful to directly relate the radial acceleration of an object to its velocity, let’s construct a direct relationship between these two variables:

\[\begin{aligned}
&a_{r}=-R \omega^{2} \\
&a_{r}=-R\left(\frac{v}{R}\right)^{2} \\
&a_{r}=-\frac{v^{2}}{R}
\end{aligned} \nonumber\]

This relationship, although mathematically equivalent to \( a_{r}=-R \omega^{2}\), is in a more “useful” form.

Another Circular Motion Scenario

A 950 kg car, traveling at a constant 30 m/s, safely makes a lefthand-turn with radius of curvature 75 m.

First, let's draw a pair of free-body diagrams for the car, a side-view (on the left) and a rearview (on the right)

The free-body diagram on the left is a side-view of the car. Notice that the upward direction is the y-direction, and the forward direction, tangent to the turn, is the t-direction.

The free-body diagram on the right is a rear-view of the car. This is what you would see if you stood directly behind the car. Notice that the upward direction is still the y-direction, and the horizontal direction, perpendicular to the direction of travel and hence directed radially outward, is the r-direction. The frictional force indicted is perpendicular to the tread on the tire. This force causes the car to accelerate toward the center of the turn. Remember, if the car is going to travel along a circular path, it must have an acceleration directed toward the center of the circle. Something has to be supplying the force that creates this acceleration. This something is the static friction between the tire and the road that acts to prevent the car from sliding out of the turn. Since the car has no velocity in the radial direction, the frictional force that points in this direction must be static!

Now that we have all that straightened out (maybe), let's apply Newton's Second Law.

\[\begin{array}{ll} \underline{\mathbf{y-direction }} && \underline{\mathbf { r-direction }} \\ \Sigma F=m a && \Sigma F=m a \\ F_{\text {road }}-F_{\text {gravity }}=m a_{y} && -F_{\text {staticfiction }}=m a_{r} \\ F_{\text {road }}-950(9.8)=950(0) && -F_{\text {staticfiction }}=m\left(-\frac{v^{2}}{R}\right) \\ F_{\text {road }}=9310 N && F_{\text {staticfriction }}=950\left(\frac{30^{2}}{75}\right) \\ && F_{\text {staticfriction }}=11400 \mathrm{~N} \end{array} \nonumber\]

Thus, to safely make this turn requires at least 11400 N of static friction. Using this value, I should be able to compute the minimum coefficient of friction necessary for the car to safely round this turn at this speed.

\[\begin{aligned}
&F_{\text {staticfriction }} \leq \mu_{s} F_{\text {road }} \\
&11400 \leq \mu_{s}(9310) \\
&\mu_{s} \geq 1.22
\end{aligned} \nonumber\]

Although this is a large value for the coefficient of static friction, it is an attainable value for a sports car with performance tires.

Activities

Six artificial satellites of mass, m, circle a space station at constant speed, v.

a. If the distance between the space station and the satellites is the same for all satellites, rank the magnitude of the net force acting on each satellite.