Kinetics of Particles: Force, Mass and Acceleration

Polar coordinates $(r,\theta)$ locate a particle by its distance $r$ from the pole O and the angle $\theta$ of the radius vector from a fixed line. Two unit vectors are used: $\mathbf{e}_r$ along the radius vector and $\mathbf{e}_\theta$ perpendicular to it (in the direction of increasing $\theta$).

          y
          |      P
          |    / |
          |  /   | e_theta
          | / r  |-> e_r
          |/ th
     O----+------------ x

Rate of change of unit vectors

Because the directions of $\mathbf{e}_r$ and $\mathbf{e}_\theta$ rotate with $\theta$:

$$\dot{\mathbf{e}}_r = \dot\theta\,\mathbf{e}_\theta \qquad \dot{\mathbf{e}}_\theta = -\dot\theta\,\mathbf{e}_r$$

Velocity

The position vector is $\mathbf{r} = r\,\mathbf{e}_r$. Differentiating:

$$\mathbf{v} = \dot r\,\mathbf{e}_r + r\dot{\mathbf{e}}_r = \dot r\,\mathbf{e}_r + r\dot\theta\,\mathbf{e}_\theta$$

So $v_r = \dot r$ (radial) and $v_\theta = r\dot\theta$ (transverse), with $v = \sqrt{v_r^2 + v_\theta^2}$.

Acceleration

Differentiating again:

$$\begin{aligned} \mathbf{a} &= \ddot r\,\mathbf{e}_r + \dot r\dot\theta\,\mathbf{e}_\theta + (\dot r\dot\theta + r\ddot\theta)\,\mathbf{e}_\theta - r\dot\theta^2\,\mathbf{e}_r \\ &= (\ddot r - r\dot\theta^2)\,\mathbf{e}_r + (r\ddot\theta + 2\dot r\dot\theta)\,\mathbf{e}_\theta \end{aligned}$$ $$a_r = \ddot r - r\dot\theta^2 \qquad a_\theta = r\ddot\theta + 2\dot r\dot\theta$$

The term $2\dot r\dot\theta$ is the Coriolis component; $r\dot\theta^2$ is the centripetal term.

Equations of motion

Applying Newton's second law $\Sigma \mathbf{F} = m\mathbf{a}$ along the two directions:

$$\Sigma F_r = m(\ddot r - r\dot\theta^2) \qquad \Sigma F_\theta = m(r\ddot\theta + 2\dot r\dot\theta)$$

For a central force ($\Sigma F_\theta = 0$) the second equation gives $\dfrac{d}{dt}(r^2\dot\theta) = 0$, i.e. constant angular momentum $h = r^2\dot\theta$ (Kepler's law of areas).

Given: $m_A = 40$ kg, $m_B = 25$ kg, $\mu = 0.20$, $g = 9.81$ m/s². The string is inextensible, so both blocks have the same acceleration $a$. B moves down and A moves toward the pulley.

   +-----+        ___
   |  A  |-------(   )     pulley
 ==+-----+==========\ \====
   friction <--      \ \ T
                      [B] (down, a)

Free-body equations

For A (horizontal): friction $F = \mu m_A g = 0.20 \times 40 \times 9.81 = 78.48$ N.

$$T - 78.48 = 40a$$

For B (vertical): $m_B g = 25 \times 9.81 = 245.25$ N.

$$245.25 - T = 25a$$

(a) Acceleration

Adding the two equations:

$$245.25 - 78.48 = 65a \;\Rightarrow\; a = \frac{166.77}{65} = 2.566\ \text{m/s}^2$$

(b) Tension

$$T = 78.48 + 40(2.566) = 181.1\ \text{N}$$

Check from B: $T = 25(9.81 - 2.566) = 181.1$ N. It agrees.

(c) Speed and distance after 2 s

Starting from rest with constant $a$:

$$v = at = 2.566 \times 2 = 5.13\ \text{m/s} \qquad s = \tfrac12 a t^2 = 5.13\ \text{m}$$

(d) Load on the pulley support

The string pulls on the pulley with $T$ horizontally (toward A) and $T$ vertically (toward B). The pulley is light and smooth, so the support must supply the resultant:

$$R = \sqrt{T^2 + T^2} = \sqrt{2}\,T = 256.1\ \text{N}$$

inclined at $45^\circ$ below the horizontal toward the B side.

Answer: a = 2.57 m/s²; T = 181.1 N; v = 5.13 m/s and s = 5.13 m at 2 s; pulley load = 256.1 N.

(a) Dynamic equilibrium and inertia force

Inertia force is the fictitious force, equal to the mass times acceleration and directed opposite to the acceleration of the body:

$$\mathbf{F}_i = -m\mathbf{a}$$

D'Alembert's principle states that if the inertia force is added to the real forces acting on a body, the body can be treated as being in (dynamic) equilibrium, so that

$$\Sigma \mathbf{F} - m\mathbf{a} = 0 \quad (\text{i.e. } \Sigma\mathbf{F} + \mathbf{F}_i = 0)$$

A dynamics problem is thus turned into a statics problem, and the equations of statics (including moment equations) can be used. For a rotating body the inertia couple $-I\alpha$ is also added.

(b) Scale reading in the lift

Forces on the man: weight $mg = 75 \times 9.81 = 735.75$ N downward and scale reaction $R$ upward. Take upward positive and let $a$ be the lift's upward acceleration; then $R - mg = ma$.

Cable tension (lift + man = 575 kg)

Same equation for the whole system: $T - Mg = Ma$, so $T = M(g + a)$.

• (i) $T = 575(9.81 + 1.5) = 6503.2$ N
• (ii) $T = 575(9.81 - 2) = 4490.8$ N

Answer: Scale reading = 848.2 N (i), 585.8 N (ii), 0 (iii); cable tension = 6503 N (i), 4491 N (ii).

Given: $r = 2 + 0.5t^2$, $\theta = 0.4t^2$, at $t = 2$ s.

Derivatives at t = 2 s

Velocity

$$\begin{aligned} v_r &= \dot r = 2.00\ \text{m/s} \\ v_\theta &= r\dot\theta = 4 \times 1.6 = 6.40\ \text{m/s} \\ v &= \sqrt{2^2 + 6.4^2} = 6.71\ \text{m/s} \end{aligned}$$

Angle from the radial line: $\tan\phi = v_\theta / v_r = 6.4/2$, so $\phi = 72.6^\circ$.

Acceleration

$$\begin{aligned} a_r &= \ddot r - r\dot\theta^2 = 1 - 4(1.6)^2 = -9.24\ \text{m/s}^2 \\ a_\theta &= r\ddot\theta + 2\dot r\dot\theta = 4(0.8) + 2(2)(1.6) = 9.60\ \text{m/s}^2 \\ a &= \sqrt{(-9.24)^2 + 9.6^2} = 13.32\ \text{m/s}^2 \end{aligned}$$

Since $a_r$ is negative and $a_\theta$ positive, the acceleration lies in the second quadrant of the $(e_r, e_\theta)$ axes, at $133.9^\circ$ measured from the $+e_r$ direction toward $+e_\theta$ (that is, $46.1^\circ$ from the $-e_r$ direction).

Answer: v = 6.71 m/s at 72.6° to the radial line; a = 13.32 m/s² at 133.9° to the radial line.