3.03q - Variable forces and vector dynamics

3.03q - Variable forces and vector dynamics

A changing force needs a changing-acceleration model. You will combine vector force equations with calculus to recover velocity and position, use conditions at the correct times, and distinguish where a particle is moving from the direction in which its velocity is changing.

From a time-dependent resultant to motion

Must a particle travel in the direction of its resultant force? Consider a force whose horizontal component stays fixed while its vertical component grows with time. The acceleration turns towards the vertical, but the velocity also remembers how the particle was moving before. Calculus connects these two pieces of information.

You already know

For fixed mass, F=ma\mathbf F=m\mathbf a, where F\mathbf F is the resultant. Differentiate position for velocity and velocity for acceleration; reverse those steps by integration. Work component by component.

A coefficient in a time-dependent expression carries whatever units make the complete component a force when time is measured in seconds. Keep a separate integration constant for each component.

Motion from a time-dependent resultant

For constant mass mm, with F(t)\mathbf F(t) the complete resultant,

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

Use velocity and position conditions to determine the constants in the last two lines. The constant-acceleration equations do not apply over an interval with varying acceleration.

Not in the formula sheet: learn it.

Here the supplied force is explicitly the resultant, so no further forces need adding.

Worked example 1: two vector integrations

Question. A 2 kg2\,\mathrm{kg} particle has resultant 4i+6tj N4\mathbf i+6t\mathbf j\,\mathrm N, for 0≤t≤20\le t\le2, with tt in seconds. Initially v=i−2j m s−1\mathbf v=\mathbf i-2\mathbf j\,\mathrm{m\,s^{-1}} and x=−i+3j m\mathbf x=-\mathbf i+3\mathbf j\,\mathrm m. Find velocity and position functions, and their values at t=2t=2, exactly.

Plan. Divide by mass, integrate twice, and use each initial condition at its own stage.

a=2i+3tjresultant divided by massv=(2t+Cx)i+(32t2+Cy)jintegrate accelerationCx=1,Cy=−2initial velocityx=(t2+t+Dx)i+(12t3−2t+Dy)jintegrate velocityDx=−1,Dy=3initial position\begin{aligned} \mathbf a&=2\mathbf i+3t\mathbf j &&\text{resultant divided by mass} \\ \mathbf v&=(2t+C_x)\mathbf i+\left(\frac32t^2+C_y\right)\mathbf j &&\text{integrate acceleration} \\ C_x&=1,\quad C_y=-2 &&\text{initial velocity} \\ \mathbf x&=(t^2+t+D_x)\mathbf i+\left(\frac12t^3-2t+D_y\right)\mathbf j &&\text{integrate velocity} \\ D_x&=-1,\quad D_y=3 &&\text{initial position} \end{aligned}

Why two pairs of constants? Initial velocity and initial position are independent information; one does not determine the other.

Answer.

v(t)=(2t+1)i+(32t2−2)jx(t)=(t2+t−1)i+(12t3−2t+3)jv(2)=5i+4j m s−1x(2)=5i+3j m\begin{aligned} \mathbf v(t)&=(2t+1)\mathbf i+\left(\frac32t^2-2\right)\mathbf j \\ \mathbf x(t)&=(t^2+t-1)\mathbf i+\left(\frac12t^3-2t+3\right)\mathbf j \\ \mathbf v(2)&=5\mathbf i+4\mathbf j\,\mathrm{m\,s^{-1}} \\ \mathbf x(2)&=5\mathbf i+3\mathbf j\,\mathrm m \end{aligned}

Check. Differentiating x\mathbf x twice gives 2i+3tj2\mathbf i+3t\mathbf j; substituting zero recovers both initial conditions.

Change only the initial horizontal velocity. The acceleration remains exactly the same.

Worked example 2: a different initial velocity

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

Plan. Integrate the same acceleration and apply the changed velocity condition.

a=2i+3tjunchanged forcev=(2t−1)i+(32t2−2)jnew horizontal constantx=(t2−t−1)i+(12t3−2t+3)jsame initial positionv(2)=3i+4jsubstitute timex(2)=i+3jsubstitute time\begin{aligned} \mathbf a&=2\mathbf i+3t\mathbf j &&\text{unchanged force} \\ \mathbf v&=(2t-1)\mathbf i+\left(\frac32t^2-2\right)\mathbf j &&\text{new horizontal constant} \\ \mathbf x&=(t^2-t-1)\mathbf i+\left(\frac12t^3-2t+3\right)\mathbf j &&\text{same initial position} \\ \mathbf v(2)&=3\mathbf i+4\mathbf j &&\text{substitute time} \\ \mathbf x(2)&=\mathbf i+3\mathbf j &&\text{substitute time} \end{aligned}

Why this step? A velocity change contributes a linear-in-time change to position, even when the force is unchanged.

Answer. The functions are as displayed; at 2 s2\,\mathrm s, velocity is 3i+4j m s−13\mathbf i+4\mathbf j\,\mathrm{m\,s^{-1}} and position is i+3j m\mathbf i+3\mathbf j\,\mathrm m.

Check. Compared with example 1, horizontal velocity is smaller by 2 m s−12\,\mathrm{m\,s^{-1}}, so horizontal position at 2 s2\,\mathrm s is smaller by 4 m4\,\mathrm m.

Position and displacement are different. Example 1 starts at −i+3j-\mathbf i+3\mathbf j and ends at 5i+3j5\mathbf i+3\mathbf j, so its displacement is 6i m6\mathbf i\,\mathrm m. If another particle has simultaneous position i−2j m\mathbf i-2\mathbf j\,\mathrm m, its position relative to the first is −4i−5j m-4\mathbf i-5\mathbf j\,\mathrm m, with separation 41 m\sqrt{41}\,\mathrm m. Subtract simultaneous positions in the requested order; the relative-position subtraction and magnitude rules are not in the formula sheet.

Watch out

Wrong: use x=12a(t)t2\mathbf x=\tfrac12\mathbf a(t)t^2 for variable acceleration. Correct: integrate the time-dependent acceleration, applying velocity and position conditions separately.

Now let the horizontal, rather than vertical, force component vary.

Conditions at a specified time

An integration constant is not automatically the value supplied in a question: that shortcut works only when the time-dependent terms vanish at the condition time.

If velocity is given at t=1t=1, substitute t=1t=1 into the entire velocity expression. An equivalent method uses definite integrals from the actual condition time. For example, v(t)=v(1)+∫1ta(u) du\mathbf v(t)=\mathbf v(1)+\int_1^t\mathbf a(u)\,\mathrm du; the dummy variable uu avoids confusing the integration variable with its upper limit. This calculus relationship is not in the formula sheet.

Worked example 3: conditions at one second

Question. A 3 kg3\,\mathrm{kg} particle has resultant 6ti−6j N6t\mathbf i-6\mathbf j\,\mathrm N, for 0≤t≤30\le t\le3, with tt in seconds. At t=1t=1, 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. Find the position function and position at t=3t=3, exactly.

Plan. Apply both conditions at one second after the appropriate integration.

a=2ti−2jdivide by massv=(t2+Cx)i+(−2t+Cy)jintegrateCx=1,Cy=5use velocity at t=1x=(t33+t+Dx)i+(5t−t2+Dy)jintegrate againDx=−103,Dy=0use position at t=1\begin{aligned} \mathbf a&=2t\mathbf i-2\mathbf j &&\text{divide by mass} \\ \mathbf v&=(t^2+C_x)\mathbf i+(-2t+C_y)\mathbf j &&\text{integrate} \\ C_x&=1,\quad C_y=5 &&\text{use velocity at }t=1 \\ \mathbf x&=\left(\frac{t^3}{3}+t+D_x\right)\mathbf i+(5t-t^2+D_y)\mathbf j &&\text{integrate again} \\ D_x&=-\frac{10}{3},\quad D_y=0 &&\text{use position at }t=1 \end{aligned}

Why substitute the actual time? The supplied velocity includes acceleration already accumulated by that time.

Answer.

x(t)=(t33+t−103)i+(5t−t2)j m,x(3)=263i+6j m\mathbf x(t)=\left(\frac{t^3}{3}+t-\frac{10}{3}\right)\mathbf i+(5t-t^2)\mathbf j\,\mathrm m, \qquad \mathbf x(3)=\frac{26}{3}\mathbf i+6\mathbf j\,\mathrm m

Check. At t=1t=1, position is −2i+4j-2\mathbf i+4\mathbf j; at t=2t=2 it is 43i+6j\tfrac43\mathbf i+6\mathbf j.

Change only the time at which those same velocity and position vectors are given. The force law and requested time stay fixed.

Worked example 4: conditions at two seconds

Question. For the same mass and force, the given velocity 2i+3j m s−12\mathbf i+3\mathbf j\,\mathrm{m\,s^{-1}} and position −2i+4j m-2\mathbf i+4\mathbf j\,\mathrm m now occur at t=2t=2. Find the position function and position at t=3t=3, exactly.

Plan. Integrate the same acceleration, but determine every constant at two seconds.

v=(t2+Cx)i+(−2t+Cy)jsame antiderivativeCx=2−4=−2,Cy=3+4=7velocity at t=2x=(t33−2t+Dx)i+(7t−t2+Dy)jintegrate velocityDx=−23,Dy=−6position at t=2\begin{aligned} \mathbf v&=(t^2+C_x)\mathbf i+(-2t+C_y)\mathbf j &&\text{same antiderivative} \\ C_x&=2-4=-2,\quad C_y=3+4=7 &&\text{velocity at }t=2 \\ \mathbf x&=\left(\frac{t^3}{3}-2t+D_x\right)\mathbf i+(7t-t^2+D_y)\mathbf j &&\text{integrate velocity} \\ D_x&=-\frac23,\quad D_y=-6 &&\text{position at }t=2 \end{aligned}

Why this step? Moving the condition time changes the entire motion satisfying the same force law.

Answer.

x(t)=(t33−2t−23)i+(7t−t2−6)j m,x(3)=73i+6j m\mathbf x(t)=\left(\frac{t^3}{3}-2t-\frac23\right)\mathbf i+(7t-t^2-6)\mathbf j\,\mathrm m, \qquad \mathbf x(3)=\frac73\mathbf i+6\mathbf j\,\mathrm m

Check. Substituting t=2t=2 gives −2i+4j-2\mathbf i+4\mathbf j, and differentiating there gives 2i+3j2\mathbf i+3\mathbf j.

Conditions at non-zero times are not a new integration rule; they require careful substitution into the familiar rule.

Watch out

Wrong: set Cx=2C_x=2 because the supplied horizontal velocity is 22. Correct: in example 4, solve 4+Cx=24+C_x=2, giving Cx=−2C_x=-2.

Apply this to another pair of conditions.

Other forces and different directions

The velocity vector gives the direction of motion; the resultant and acceleration give the direction in which velocity is changing.

Continue this lesson.

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