BE Civil Engineering (IOE, TU) Applied Mechanics - Dynamics (IOE, CE 452) Question Paper 2076 Nepal

Set-up. Take the launch point as origin, $x$ horizontal (positive in direction of throw), $y$ vertical (positive up). The ground is at $y = -45\ \text{m}$.

Initial velocity components:

$$u_x = u\cos\theta = 25\cos 35^\circ = 25(0.8192) = 20.48\ \text{m/s}$$ $$u_y = u\sin\theta = 25\sin 35^\circ = 25(0.5736) = 14.34\ \text{m/s}$$

(a) Time of flight. Vertical motion: $y = u_y t - \tfrac{1}{2}g t^2$. At impact $y = -45$:

$$-45 = 14.34\,t - 4.905\,t^2$$ $$4.905\,t^2 - 14.34\,t - 45 = 0$$

Using the quadratic formula:

$$t = \frac{14.34 \pm \sqrt{14.34^2 + 4(4.905)(45)}}{2(4.905)} = \frac{14.34 \pm \sqrt{205.6 + 882.9}}{9.81} = \frac{14.34 \pm 33.00}{9.81}$$

Taking the positive root: $t = \dfrac{14.34 + 33.00}{9.81} = \dfrac{47.34}{9.81}$.

Time of flight $t = 4.83\ \text{s}$.

(b) Horizontal range.

$$R = u_x\, t = 20.48 \times 4.83 = 98.9\ \text{m}$$

Range from foot of cliff $R = 98.9\ \text{m}$.

(c) Maximum height above launch point. At the apex $v_y = 0$:

$$h_{max} = \frac{u_y^2}{2g} = \frac{14.34^2}{2(9.81)} = \frac{205.6}{19.62} = 10.48\ \text{m}$$

Maximum height above launch point $= 10.48\ \text{m}$ (i.e. 55.48 m above the ground).

Velocity at impact: horizontal component unchanged, $v_x = 20.48\ \text{m/s}$. Vertical component: $v_y = u_y - g t = 14.34 - 9.81(4.83) = 14.34 - 47.38 = -33.04\ \text{m/s}$ (downward).

$$|v| = \sqrt{v_x^2 + v_y^2} = \sqrt{20.48^2 + 33.04^2} = \sqrt{419.4 + 1091.6} = \sqrt{1511.0} = 38.87\ \text{m/s}$$

Direction below horizontal:

$$\alpha = \tan^{-1}\!\left(\frac{33.04}{20.48}\right) = \tan^{-1}(1.613) = 58.2^\circ$$

Impact velocity $= 38.9\ \text{m/s}$ at $58.2^\circ$ below the horizontal.

(a) Speed at the spring (after sliding 6 m).

Forces along the incline over the 6 m slide:

• Component of weight down the plane: $W_\parallel = mg\sin 30^\circ = 8(9.81)(0.5) = 39.24\ \text{N}$
• Normal force: $N = mg\cos 30^\circ = 8(9.81)(0.8660) = 67.97\ \text{N}$
• Friction (opposes motion, acts up the plane): $f = \mu_k N = 0.25(67.97) = 16.99\ \text{N}$

Work-energy theorem from rest over $s = 6\ \text{m}$:

$$\tfrac{1}{2}m v^2 - 0 = (W_\parallel - f)\,s = (39.24 - 16.99)(6) = (22.25)(6) = 133.5\ \text{J}$$ $$v = \sqrt{\frac{2(133.5)}{8}} = \sqrt{33.38} = 5.78\ \text{m/s}$$

Speed just before contacting the spring $v = 5.78\ \text{m/s}$.

(b) Maximum spring compression $x$.

From contact to maximum compression the block moves a further distance $x$ down the plane and comes momentarily to rest. Apply work-energy from the moment of contact (speed $v$) to rest:

$$0 - \tfrac{1}{2}m v^2 = (W_\parallel - f)\,x - \tfrac{1}{2}k x^2$$

The gravity and friction net driving force still acts ($W_\parallel - f = 22.25\ \text{N}$), and the spring stores $\tfrac12 k x^2$.

$$\tfrac{1}{2}(2000)x^2 - (22.25)x - 133.5 = 0$$ $$1000\,x^2 - 22.25\,x - 133.5 = 0$$

Solve:

$$x = \frac{22.25 \pm \sqrt{22.25^2 + 4(1000)(133.5)}}{2(1000)} = \frac{22.25 \pm \sqrt{495.1 + 534000}}{2000} = \frac{22.25 \pm 731.1}{2000}$$

Taking the positive root: $x = \dfrac{22.25 + 731.1}{2000} = \dfrac{753.4}{2000} = 0.377\ \text{m}$.

Maximum spring compression $x = 0.377\ \text{m}$ (≈ 377 mm).

Sign convention: take rightward as positive. Then $u_A = +6\ \text{m/s}$, $u_B = -2\ \text{m/s}$, $m_A = 3\ \text{kg}$, $m_B = 5\ \text{kg}$.

(a) Velocities after impact.

Conservation of linear momentum:

$$m_A u_A + m_B u_B = m_A v_A + m_B v_B$$ $$3(6) + 5(-2) = 3 v_A + 5 v_B \;\Rightarrow\; 18 - 10 = 3 v_A + 5 v_B \;\Rightarrow\; 3 v_A + 5 v_B = 8 \quad (1)$$

Coefficient of restitution:

$$e = \frac{v_B - v_A}{u_A - u_B} \;\Rightarrow\; 0.7 = \frac{v_B - v_A}{6 - (-2)} = \frac{v_B - v_A}{8}$$ $$v_B - v_A = 5.6 \quad (2)$$

From (2): $v_B = v_A + 5.6$. Substitute into (1):

$$3 v_A + 5(v_A + 5.6) = 8 \;\Rightarrow\; 8 v_A + 28 = 8 \;\Rightarrow\; 8 v_A = -20 \;\Rightarrow\; v_A = -2.5\ \text{m/s}$$ $$v_B = -2.5 + 5.6 = 3.1\ \text{m/s}$$

After impact: $v_A = 2.5\ \text{m/s}$ to the LEFT, $v_B = 3.1\ \text{m/s}$ to the RIGHT.

(b) Kinetic energy loss.

Before:

$$KE_i = \tfrac12(3)(6^2) + \tfrac12(5)(2^2) = 54 + 10 = 64\ \text{J}$$

After:

$$KE_f = \tfrac12(3)(2.5^2) + \tfrac12(5)(3.1^2) = \tfrac12(3)(6.25) + \tfrac12(5)(9.61) = 9.375 + 24.025 = 33.40\ \text{J}$$ $$\Delta KE = KE_i - KE_f = 64 - 33.40 = 30.6\ \text{J}$$

Loss of kinetic energy $\approx 30.6\ \text{J}$.

Convert peak speed to rad/s:

$$\omega = 300\ \text{rpm} = 300 \times \frac{2\pi}{60} = 31.42\ \text{rad/s}$$

(a) $\omega$–$t$ diagram (description). A trapezoid: a straight line rising from $(0,0)$ to $(12\,\text{s}, 31.42\,\text{rad/s})$ [acceleration], a horizontal line from $t=12$ to $t=32\,\text{s}$ at $31.42\,\text{rad/s}$ [constant], then a straight line falling from $(32,31.42)$ to $(40,0)$ [deceleration].

omega (rad/s)
31.42 |      ________________
      |     /                \
      |    /                  \
      |   /                    \
    0 |__/______________________\____ t (s)
        0  12               32  40

(b) Angular acceleration in each phase.

Phase 1 (acceleration, $0 \to 12\,\text{s}$):

$$\alpha_1 = \frac{31.42 - 0}{12} = 2.618\ \text{rad/s}^2$$

Phase 2 (constant speed, $12 \to 32\,\text{s}$):

$$\alpha_2 = 0\ \text{rad/s}^2$$

Phase 3 (braking, $32 \to 40\,\text{s}$):

$$\alpha_3 = \frac{0 - 31.42}{8} = -3.927\ \text{rad/s}^2$$

(c) Total revolutions = total angle / $2\pi$.

Angle in each phase equals area under the $\omega$–$t$ graph.

Phase 1: $\theta_1 = \tfrac12 (12)(31.42) = 188.5\ \text{rad}$

Phase 2: $\theta_2 = (31.42)(20) = 628.3\ \text{rad}$

Phase 3: $\theta_3 = \tfrac12 (8)(31.42) = 125.7\ \text{rad}$

Total angle:

$$\theta = 188.5 + 628.3 + 125.7 = 942.5\ \text{rad}$$ $$N = \frac{942.5}{2\pi} = \frac{942.5}{6.2832} = 150.0\ \text{rev}$$

Total number of revolutions $\approx 150$.

Given: $m = 4\ \text{kg}$, $k = 1600\ \text{N/m}$, $c = 40\ \text{N·s/m}$.

(a) Undamped natural frequency.

$$\omega_n = \sqrt{\frac{k}{m}} = \sqrt{\frac{1600}{4}} = \sqrt{400} = 20\ \text{rad/s}$$ $$f_n = \frac{\omega_n}{2\pi} = \frac{20}{6.2832} = 3.18\ \text{Hz}$$

Undamped natural frequency $\omega_n = 20\ \text{rad/s}$ ($f_n = 3.18\ \text{Hz}$).

(b) Damping ratio and classification.

Critical damping coefficient:

$$c_c = 2\sqrt{km} = 2\sqrt{1600 \times 4} = 2\sqrt{6400} = 2(80) = 160\ \text{N·s/m}$$ $$\zeta = \frac{c}{c_c} = \frac{40}{160} = 0.25$$

Since $\zeta = 0.25 < 1$, the system is underdamped.

Damped natural frequency:

$$\omega_d = \omega_n\sqrt{1-\zeta^2} = 20\sqrt{1-0.0625} = 20\sqrt{0.9375} = 20(0.9682) = 19.36\ \text{rad/s}$$

$\zeta = 0.25$ (underdamped); $\omega_d = 19.36\ \text{rad/s}$.

(c) Logarithmic decrement and amplitude after 3 cycles.

Logarithmic decrement:

$$\delta = \frac{2\pi\zeta}{\sqrt{1-\zeta^2}} = \frac{2\pi(0.25)}{0.9682} = \frac{1.5708}{0.9682} = 1.622$$

The amplitude ratio over $n$ cycles obeys $\dfrac{x_0}{x_n} = e^{n\delta}$. For $n = 3$:

$$\frac{x_0}{x_3} = e^{3(1.622)} = e^{4.866} = 129.9$$ $$x_3 = \frac{x_0}{129.9} = \frac{30\ \text{mm}}{129.9} = 0.231\ \text{mm}$$

Logarithmic decrement $\delta = 1.62$; amplitude after 3 cycles $x_3 \approx 0.231\ \text{mm}$.

Normal (centripetal) component:

$$a_n = \frac{v^2}{\rho} = \frac{20^2}{200} = \frac{400}{200} = 2.0\ \text{m/s}^2$$

Tangential component (given, rate of change of speed):

$$a_t = 3.0\ \text{m/s}^2$$

Resultant acceleration magnitude:

$$a = \sqrt{a_t^2 + a_n^2} = \sqrt{3.0^2 + 2.0^2} = \sqrt{9 + 4} = \sqrt{13} = 3.61\ \text{m/s}^2$$

Angle with direction of motion (the tangent):

$$\phi = \tan^{-1}\!\left(\frac{a_n}{a_t}\right) = \tan^{-1}\!\left(\frac{2.0}{3.0}\right) = \tan^{-1}(0.667) = 33.7^\circ$$

Total acceleration $= 3.61\ \text{m/s}^2$ at $33.7^\circ$ from the direction of motion (toward the centre of the curve).

Free-body equations. Let $a$ be the acceleration; the heavier block A descends, B rises. Let $T$ be the string tension.

For block A (descending):

$$m_A g - T = m_A a$$

For block B (ascending):

$$T - m_B g = m_B a$$

Add the two equations:

$$(m_A - m_B)g = (m_A + m_B)a$$ $$a = \frac{(m_A - m_B)g}{m_A + m_B} = \frac{(10 - 6)(9.81)}{10 + 6} = \frac{4(9.81)}{16} = \frac{39.24}{16} = 2.453\ \text{m/s}^2$$

Acceleration $a = 2.45\ \text{m/s}^2$.

Tension (from block B's equation):

$$T = m_B(g + a) = 6(9.81 + 2.453) = 6(12.263) = 73.58\ \text{N}$$

Check with block A: $T = m_A(g - a) = 10(9.81 - 2.453) = 10(7.357) = 73.57\ \text{N}$ ✓

Tension $T = 73.6\ \text{N}$.

Rolling without slipping: the instantaneous centre of rotation (ICR) is the contact point $C$ with the ground (its velocity is zero).

(a) Angular velocity. The centre $O$ is at distance $r = 0.4\ \text{m}$ from $C$ and moves at $v_O = 5\ \text{m/s}$:

$$\omega = \frac{v_O}{r} = \frac{5}{0.4} = 12.5\ \text{rad/s}$$

Angular velocity $\omega = 12.5\ \text{rad/s}$.

(b) Top point velocity. The top point $T$ is at distance $2r = 0.8\ \text{m}$ from the ICR:

$$v_T = \omega(2r) = 12.5 \times 0.8 = 10\ \text{m/s}\ \text{(horizontal, forward)}$$

Velocity of top point $= 10\ \text{m/s}$.

(c) Leading horizontal rim point (level with centre, ahead of it). Its distance from the ICR is the diagonal:

$$d = \sqrt{r^2 + r^2} = r\sqrt{2} = 0.4\sqrt{2} = 0.5657\ \text{m}$$ $$v = \omega\,d = 12.5 \times 0.5657 = 7.07\ \text{m/s}$$

Its direction is perpendicular to the line joining it to $C$, i.e. at $45^\circ$ above the horizontal (directed forward and upward).

Velocity of the side rim point $= 7.07\ \text{m/s}$ at $45^\circ$ above horizontal.

(a) Mass moment of inertia of a solid cylinder about its central axis:

$$I = \tfrac12 m r^2 = \tfrac12 (20)(0.3^2) = \tfrac12 (20)(0.09) = 0.90\ \text{kg·m}^2$$

$I = 0.90\ \text{kg·m}^2$.

(b) Angular acceleration from $M = I\alpha$, where the applied torque is $M = F r$:

$$M = F r = 50 \times 0.3 = 15\ \text{N·m}$$ $$\alpha = \frac{M}{I} = \frac{15}{0.90} = 16.67\ \text{rad/s}^2$$

$\alpha = 16.67\ \text{rad/s}^2$.

(c) Angular velocity after 4 s (from rest, constant $\alpha$):

$$\omega = \omega_0 + \alpha t = 0 + 16.67(4) = 66.67\ \text{rad/s}$$

$\omega = 66.7\ \text{rad/s}$ after 4 s (equivalently $66.67 \times 60/2\pi = 636.6\ \text{rpm}$).

Convert initial speed:

$$\omega_0 = 240\ \text{rpm} = 240 \times \frac{2\pi}{60} = 25.13\ \text{rad/s}$$

Angular impulse-momentum principle: $M\,t = I(\omega_f - \omega_0)$. With $\omega_f = 0$:

$$M(15) = 12(0 - 25.13) = -301.6\ \text{kg·m}^2/\text{s}$$ $$M = \frac{-301.6}{15} = -20.11\ \text{N·m}$$

The negative sign indicates a decelerating (braking) torque.

Braking torque magnitude $= 20.1\ \text{N·m}$.

Revolutions during braking. Angular deceleration:

$$\alpha = \frac{\omega_f - \omega_0}{t} = \frac{0 - 25.13}{15} = -1.675\ \text{rad/s}^2$$

Angle turned (using $\omega_f^2 = \omega_0^2 + 2\alpha\theta$):

$$0 = 25.13^2 + 2(-1.675)\theta \;\Rightarrow\; \theta = \frac{631.5}{3.351} = 188.5\ \text{rad}$$ $$N = \frac{188.5}{2\pi} = \frac{188.5}{6.2832} = 30.0\ \text{rev}$$

Number of revolutions during braking $\approx 30$.

(a) Resonance. Resonance occurs in a forced vibration when the forcing (excitation) frequency $\omega$ approaches the natural frequency $\omega_n$ of the system, i.e. the frequency ratio $r = \omega/\omega_n \to 1$. At resonance the magnification factor becomes very large (theoretically infinite for an undamped system), so even a small periodic force produces very large amplitudes of vibration.

Engineering significance: In civil structures (bridges, tall buildings, floors, machine foundations) resonance can cause excessive deflections, fatigue, discomfort and even collapse. Designers must ensure the natural frequencies of a structure are well separated from the dominant frequencies of expected dynamic loads (wind, earthquake, pedestrian/traffic, rotating machinery), and provide adequate damping.

(b) Magnification factor and steady-state amplitude.

Frequency ratio:

$$r = \frac{\omega}{\omega_n} = \frac{6}{8} = 0.75$$

For an undamped forced system the magnification factor (dynamic magnification) is:

$$MF = \frac{1}{|1 - r^2|} = \frac{1}{|1 - 0.75^2|} = \frac{1}{1 - 0.5625} = \frac{1}{0.4375} = 2.286$$

Static deflection $x_{st} = F_0/k = 5\ \text{mm}$. Steady-state amplitude:

$$X = MF \times x_{st} = 2.286 \times 5 = 11.43\ \text{mm}$$

Magnification factor $= 2.29$; steady-state amplitude $X = 11.43\ \text{mm}$.