Mn7, Mk12 - Variable forces and vector dynamics

Mn7, Mk12 - Variable forces and vector dynamics

A vector equation of motion describes how a resultant changes a particle's velocity. You will integrate time-dependent forces, use conditions at specified times, find relative positions at a common instant, and recover missing forces from prescribed motion.

From a time-dependent resultant to motion

If the horizontal resultant is zero, must horizontal motion stop? A particle already moving left does not lose that velocity just because its horizontal acceleration is momentarily zero. The force controls the change of velocity; an initial condition tells you the velocity it changes from.

You already know

For fixed mass, F=ma\mathbf F=m\mathbf a, with F\mathbf F the resultant. MEI uses r\mathbf r for position, with v=dr/dt\mathbf v=\mathrm d\mathbf r/\mathrm dt and a=dv/dt\mathbf a=\mathrm d\mathbf v/\mathrm dt. Integrate in the reverse order to recover motion.

Treat the horizontal and vertical components as two linked calculus calculations. The same time variable appears in both, but each integration needs its own constant in each component.

Time-dependent vector dynamics

For constant mass mm,

a(t)=F(t)mv(t)=∫a(t) dtr(t)=∫v(t) dt\begin{aligned} \mathbf a(t)&=\frac{\mathbf F(t)}{m} \\ \mathbf v(t)&=\int\mathbf a(t)\,\mathrm dt \\ \mathbf r(t)&=\int\mathbf v(t)\,\mathrm dt \end{aligned}

Here F\mathbf F must include every force. Use velocity and position conditions to fix the constants; do not use constant-acceleration equations across a variable-acceleration interval.

Not in the formula sheet: learn it.

The following force is already the resultant. Its coefficients have the units required when tt is measured in seconds.

Worked example 1: integrating the equation of motion

Question. A 4 kg4\,\mathrm{kg} particle has resultant 8ti−8j N8t\mathbf i-8\mathbf j\,\mathrm N, for 0≤t≤20\le t\le2. Initially v=−i+3j m s−1\mathbf v=-\mathbf i+3\mathbf j\,\mathrm{m\,s^{-1}} and r=2i−4j m\mathbf r=2\mathbf i-4\mathbf j\,\mathrm m. Find velocity and position functions, and their values at t=2t=2, exactly.

Plan. Divide by mass, integrate acceleration, then integrate velocity, applying both initial conditions.

a=2ti−2jdivide resultant by massv=(t2+Cx)i+(−2t+Cy)jintegrate onceCx=−1,Cy=3initial velocityr=(t33−t+Dx)i+(3t−t2+Dy)jintegrate againDx=2,Dy=−4initial position\begin{aligned} \mathbf a&=2t\mathbf i-2\mathbf j &&\text{divide resultant by mass} \\ \mathbf v&=(t^2+C_x)\mathbf i+(-2t+C_y)\mathbf j &&\text{integrate once} \\ C_x&=-1,\quad C_y=3 &&\text{initial velocity} \\ \mathbf r&=\left(\frac{t^3}{3}-t+D_x\right)\mathbf i+(3t-t^2+D_y)\mathbf j &&\text{integrate again} \\ D_x&=2,\quad D_y=-4 &&\text{initial position} \end{aligned}

Why two conditions? Velocity determines the first constants; position determines the next constants. They describe different information.

Answer.

v=(t2−1)i+(3−2t)jr=(t33−t+2)i+(3t−t2−4)jv(2)=3i−j m s−1r(2)=83i−2j m\begin{aligned} \mathbf v&=(t^2-1)\mathbf i+(3-2t)\mathbf j \\ \mathbf r&=\left(\frac{t^3}{3}-t+2\right)\mathbf i+(3t-t^2-4)\mathbf j \\ \mathbf v(2)&=3\mathbf i-\mathbf j\,\mathrm{m\,s^{-1}} \\ \mathbf r(2)&=\frac83\mathbf i-2\mathbf j\,\mathrm m \end{aligned}

Check. Two derivatives recover 2ti−2j2t\mathbf i-2\mathbf j. At zero, horizontal force is zero but horizontal velocity is −1 m s−1-1\,\mathrm{m\,s^{-1}}: the particle is moving left.

Change only the initial horizontal velocity, from −1-1 to +1 m s−1+1\,\mathrm{m\,s^{-1}}. Keep the requested outputs exact.

Worked example 2: changing the starting motion

Question. For the same mass, resultant, interval and initial position, let initial velocity be i+3j m s−1\mathbf i+3\mathbf j\,\mathrm{m\,s^{-1}}. Find velocity and position functions, and their values at t=2t=2, exactly.

Plan. Integrate the unchanged acceleration using the altered initial velocity.

a=2ti−2jsame equation of motionv=(t2+1)i+(3−2t)jnew horizontal constantr=(t33+t+2)i+(3t−t2−4)jsame initial positionv(2)=5i−jsubstitute timer(2)=203i−2jsubstitute time\begin{aligned} \mathbf a&=2t\mathbf i-2\mathbf j &&\text{same equation of motion} \\ \mathbf v&=(t^2+1)\mathbf i+(3-2t)\mathbf j &&\text{new horizontal constant} \\ \mathbf r&=\left(\frac{t^3}{3}+t+2\right)\mathbf i+(3t-t^2-4)\mathbf j &&\text{same initial position} \\ \mathbf v(2)&=5\mathbf i-\mathbf j &&\text{substitute time} \\ \mathbf r(2)&=\frac{20}{3}\mathbf i-2\mathbf j &&\text{substitute time} \end{aligned}

Why this step? The same resultant can produce different motions because it specifies changes of velocity, not the initial velocity.

Answer. The displayed functions give v(2)=5i−j m s−1\mathbf v(2)=5\mathbf i-\mathbf j\,\mathrm{m\,s^{-1}} and r(2)=203i−2j m\mathbf r(2)=\tfrac{20}{3}\mathbf i-2\mathbf j\,\mathrm m.

Check. Horizontal velocity has increased by 2 m s−12\,\mathrm{m\,s^{-1}}, so position after 2 s2\,\mathrm s has increased by 4 m4\,\mathrm m.

Example 1 also gives a small surprise: horizontal acceleration is positive for every positive time, but horizontal velocity stays negative until t=1t=1. A positive acceleration need not mean positive velocity.

Watch out

Wrong: insert the current a(t)\mathbf a(t) into r=12at2\mathbf r=\tfrac12\mathbf a t^2. Correct: integrate the variable function and retain the initial-condition terms.

Try a new force and new initial conditions.

Conditions at a specified time

Use the actual condition time when fixing constants, and the same observation time when comparing two particles' positions.

A condition at t=1t=1 is not an initial condition at zero. You can substitute into an antiderivative or use v(t)=v(1)+∫1ta(u) du\mathbf v(t)=\mathbf v(1)+\int_1^t\mathbf a(u)\,\mathrm du, then integrate velocity from the same time. These variable-acceleration relationships are not in the formula sheet.

For simultaneous positions, the position of QQ relative to PP is rQ−rP\mathbf r_Q-\mathbf r_P. This vector points from PP to QQ; subtracting in the opposite order reverses it. Relative position is a position comparison, not a new force equation, and its subtraction rule is not in the formula sheet.

Worked example 3: conditions and a relative position

Question. A 3 kg3\,\mathrm{kg} particle PP has resultant 6ti−6j N6t\mathbf i-6\mathbf j\,\mathrm N, for 0≤t≤30\le t\le3. At t=1t=1, its velocity is 2i+3j m s−12\mathbf i+3\mathbf j\,\mathrm{m\,s^{-1}} and position is −2i+4j m-2\mathbf i+4\mathbf j\,\mathrm m. At t=3t=3, QQ has position i−2j m\mathbf i-2\mathbf j\,\mathrm m. Find PP's position function, its position at t=3t=3, and QQ's position relative to PP then, exactly.

Plan. Apply the conditions at one second, then subtract simultaneous positions at three seconds.

aP=2ti−2jdivide by massvP=(t2+1)i+(5−2t)jintegrate and use t=1rP=(t33+t−103)i+(5t−t2)jintegrate and use t=1rP(3)=263i+6jobserve both at t=3rQ(3)−rP(3)=−233i−8jtarget minus reference\begin{aligned} \mathbf a_P&=2t\mathbf i-2\mathbf j &&\text{divide by mass} \\ \mathbf v_P&=(t^2+1)\mathbf i+(5-2t)\mathbf j &&\text{integrate and use }t=1 \\ \mathbf r_P&=\left(\frac{t^3}{3}+t-\frac{10}{3}\right)\mathbf i+(5t-t^2)\mathbf j &&\text{integrate and use }t=1 \\ \mathbf r_P(3)&=\frac{26}{3}\mathbf i+6\mathbf j &&\text{observe both at }t=3 \\ \mathbf r_Q(3)-\mathbf r_P(3)&=-\frac{23}{3}\mathbf i-8\mathbf j &&\text{target minus reference} \end{aligned}

Why use a common time? Subtracting positions at different times would not describe the separation of the two particles at one instant.

Answer. PP's function and position are displayed; the relative position is −233i−8j m-\tfrac{23}{3}\mathbf i-8\mathbf j\,\mathrm m.

Check. Adding the relative vector to PP's position restores i−2j\mathbf i-2\mathbf j, the position of QQ.

Change only the time at which PP's two given conditions occur. Keep the force law, QQ's observation and all requested outputs unchanged.

Worked example 4: moving the condition time

Question. The same given velocity and position of PP now occur at t=2t=2, rather than t=1t=1. Find PP's position function, its position at t=3t=3, and QQ's position relative to PP then, exactly.

Plan. Determine new integration constants at two seconds, then repeat the simultaneous comparison.

vP=(t2+Cx)i+(−2t+Cy)jintegrate accelerationCx=−2,Cy=7velocity condition at t=2rP=(t33−2t−23)i+(7t−t2−6)jposition condition at t=2rP(3)=73i+6jsame observation timerQ(3)−rP(3)=−43i−8jsubtract simultaneous positions\begin{aligned} \mathbf v_P&=(t^2+C_x)\mathbf i+(-2t+C_y)\mathbf j &&\text{integrate acceleration} \\ C_x&=-2,\quad C_y=7 &&\text{velocity condition at }t=2 \\ \mathbf r_P&=\left(\frac{t^3}{3}-2t-\frac23\right)\mathbf i+(7t-t^2-6)\mathbf j &&\text{position condition at }t=2 \\ \mathbf r_P(3)&=\frac73\mathbf i+6\mathbf j &&\text{same observation time} \\ \mathbf r_Q(3)-\mathbf r_P(3)&=-\frac43\mathbf i-8\mathbf j &&\text{subtract simultaneous positions} \end{aligned}

Why recompute all constants? The time at which a velocity or position is attained is part of that condition.

Answer. The position function and rP(3)\mathbf r_P(3) are displayed; QQ's position relative to PP is −43i−8j m-\tfrac43\mathbf i-8\mathbf j\,\mathrm m.

Check. At t=2t=2, the position function gives −2i+4j-2\mathbf i+4\mathbf j, and its derivative gives 2i+3j2\mathbf i+3\mathbf j.

A sign error in relative position changes which particle the vector starts from, even though its magnitude is unchanged.

Watch out

Wrong: use the supplied velocity components as integration constants when the condition is at t=2t=2. Correct: substitute t=2t=2 into the entire velocity function before solving for the constants.

The next calculation uses a force whose vertical component varies instead.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account