BE Civil Engineering (IOE, TU) Applied Mechanics - Dynamics (IOE, CE 452) Question Paper 2078 Nepal

Set up coordinate axes: origin at the launch point (top of building), $x$ horizontal (direction of throw), $y$ vertical upward. Ground is at $y = -40\ \text{m}$.

Initial velocity components:

$$v_{0x} = 25\cos 30^\circ = 25(0.8660) = 21.65\ \text{m/s}$$ $$v_{0y} = 25\sin 30^\circ = 25(0.5000) = 12.50\ \text{m/s}$$

(a) Maximum height. At the apex $v_y = 0$:

$$v_y^2 = v_{0y}^2 - 2g\,\Delta y \;\Rightarrow\; 0 = 12.50^2 - 2(9.81)\Delta y$$ $$\Delta y = \frac{156.25}{19.62} = 7.96\ \text{m (above launch point)}$$

Height above ground $= 40 + 7.96 = \mathbf{47.96\ m}$.

(b) Total time of flight. Vertical displacement to ground: $y = -40\ \text{m}$.

$$y = v_{0y}t - \tfrac{1}{2}g t^2 \;\Rightarrow\; -40 = 12.50\,t - 4.905\,t^2$$ $$4.905\,t^2 - 12.50\,t - 40 = 0$$ $$t = \frac{12.50 \pm \sqrt{12.50^2 + 4(4.905)(40)}}{2(4.905)} = \frac{12.50 \pm \sqrt{156.25 + 784.8}}{9.81}$$ $$t = \frac{12.50 + \sqrt{941.05}}{9.81} = \frac{12.50 + 30.68}{9.81} = \mathbf{4.40\ s}$$

(The negative root is rejected.)

(c) Horizontal range from foot of building.

$$x = v_{0x}\,t = 21.65 \times 4.40 = \mathbf{95.3\ m}$$

(d) Velocity at impact.

$$v_x = v_{0x} = 21.65\ \text{m/s}$$ $$v_y = v_{0y} - g t = 12.50 - 9.81(4.40) = 12.50 - 43.16 = -30.66\ \text{m/s}$$

Magnitude:

$$v = \sqrt{21.65^2 + 30.66^2} = \sqrt{468.7 + 940.0} = \sqrt{1408.7} = \mathbf{37.5\ m/s}$$

Direction below horizontal:

$$\theta = \tan^{-1}\!\left(\frac{30.66}{21.65}\right) = \tan^{-1}(1.416) = \mathbf{54.8^\circ\ below\ the\ horizontal}$$

(a) Free-body diagrams.

Block $A$ (horizontal): tension $T$ (toward pulley, $+x$), friction $f = \mu_k N$ (opposing motion, $-x$), weight $m_A g$ down, normal $N$ up.

$$N = m_A g = 15 \times 9.81 = 147.15\ \text{N}, \quad f = 0.25 \times 147.15 = 36.79\ \text{N}$$

Block $B$ (vertical): weight $m_B g$ down, tension $T$ up; accelerates downward.

(b) Acceleration. Apply Newton's second law (let $a$ be common magnitude; $A$ moves right, $B$ moves down).

Block $A$: $\;T - f = m_A a$

Block $B$: $\;m_B g - T = m_B a$

Add the equations:

$$m_B g - f = (m_A + m_B)a$$ $$98.10 - 36.79 = (25)a$$ $$a = \frac{61.31}{25} = \mathbf{2.45\ m/s^2}$$

(c) Tension. From block $B$:

$$T = m_B(g - a) = 10(9.81 - 2.45) = 10(7.36) = \mathbf{73.6\ N}$$

Check with block $A$: $T = m_A a + f = 15(2.45) + 36.79 = 36.78 + 36.79 = 73.6\ \text{N}$. ✓

(d) Distance descended in 2 s (starting from rest):

$$s = \tfrac{1}{2}a t^2 = \tfrac{1}{2}(2.45)(2)^2 = \tfrac{1}{2}(2.45)(4) = \mathbf{4.90\ m}$$

(a) Work-energy principle. The total work done by all forces acting on a particle as it moves from position 1 to position 2 equals the change in its kinetic energy:

$$U_{1\to 2} = \Delta T = \tfrac{1}{2}m v_2^2 - \tfrac{1}{2}m v_1^2$$

(b) Speed at first contact with spring. From release (rest) to contact, only gravity does work over a drop $h = 1.5\ \text{m}$:

$$m g h = \tfrac{1}{2}m v^2 \;\Rightarrow\; v = \sqrt{2 g h} = \sqrt{2(9.81)(1.5)} = \sqrt{29.43} = \mathbf{5.42\ m/s}$$

(c) Maximum spring compression. At maximum compression the collar is momentarily at rest, so $v_1 = v_2 = 0$ for the full motion (release to maximum compression). During further descent through distance $x$ the collar drops an additional $x$, so gravity does work $m g x$ over this segment, while the spring does negative work $-\tfrac{1}{2}k x^2$.

Applying work-energy from release to maximum compression (total drop $= 1.5 + x$):

$$m g (1.5 + x) - \tfrac{1}{2}k x^2 = 0$$ $$20(9.81)(1.5 + x) - \tfrac{1}{2}(8000)x^2 = 0$$ $$294.3(1.5 + x) - 4000 x^2 = 0$$ $$441.45 + 294.3\,x - 4000 x^2 = 0$$ $$4000 x^2 - 294.3 x - 441.45 = 0$$

Solve:

$$x = \frac{294.3 \pm \sqrt{294.3^2 + 4(4000)(441.45)}}{2(4000)} = \frac{294.3 \pm \sqrt{86612 + 7\,063\,200}}{8000}$$ $$x = \frac{294.3 \pm \sqrt{7\,149\,812}}{8000} = \frac{294.3 + 2673.9}{8000} = \frac{2968.2}{8000} = \mathbf{0.371\ m}$$

(The negative root is rejected.)

Maximum spring compression $\approx \mathbf{0.371\ m\ (371\ mm)}$.

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

$$I = \tfrac{1}{2}m r^2 = \tfrac{1}{2}(50)(0.4)^2 = \tfrac{1}{2}(50)(0.16) = \mathbf{4.0\ kg\cdot m^2}$$

(b) Angular acceleration. Applied torque from the cord: $T_F = F r = 120 \times 0.4 = 48\ \text{N·m}$. Net torque: $\sum M = T_F - T_{\text{friction}} = 48 - 6 = 42\ \text{N·m}$.

$$\sum M = I\alpha \;\Rightarrow\; \alpha = \frac{42}{4.0} = \mathbf{10.5\ rad/s^2}$$

(c) After $t = 5\ \text{s}$ (from rest): Angular velocity:

$$\omega = \alpha t = 10.5 \times 5 = \mathbf{52.5\ rad/s}$$

Angular displacement:

$$\theta = \tfrac{1}{2}\alpha t^2 = \tfrac{1}{2}(10.5)(25) = 131.25\ \text{rad}$$

Number of revolutions:

$$N = \frac{\theta}{2\pi} = \frac{131.25}{6.2832} = \mathbf{20.9\ revolutions}$$

(a) Undamped natural frequency.

$$\omega_n = \sqrt{\frac{k}{m}} = \sqrt{\frac{12000}{30}} = \sqrt{400} = \mathbf{20\ rad/s}$$ $$f_n = \frac{\omega_n}{2\pi} = \frac{20}{6.2832} = \mathbf{3.18\ Hz}$$

(b) Damping ratio. Critical damping coefficient:

$$c_c = 2\sqrt{k m} = 2\sqrt{12000 \times 30} = 2\sqrt{360000} = 2(600) = 1200\ \text{N·s/m}$$ $$\zeta = \frac{c}{c_c} = \frac{240}{1200} = \mathbf{0.20}$$

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

(c) Damped natural frequency and period.

$$\omega_d = \omega_n\sqrt{1 - \zeta^2} = 20\sqrt{1 - 0.04} = 20\sqrt{0.96} = 20(0.9798) = \mathbf{19.60\ rad/s}$$ $$T_d = \frac{2\pi}{\omega_d} = \frac{6.2832}{19.60} = \mathbf{0.321\ s}$$

(d) Amplitude after 4 cycles (logarithmic decrement). Logarithmic decrement:

$$\delta = \frac{2\pi\zeta}{\sqrt{1-\zeta^2}} = \frac{2\pi(0.20)}{0.9798} = \frac{1.2566}{0.9798} = 1.2825$$

For $n = 4$ cycles:

$$\ln\!\left(\frac{x_0}{x_4}\right) = n\delta = 4(1.2825) = 5.130$$ $$\frac{x_0}{x_4} = e^{5.130} = 169.0$$ $$x_4 = \frac{20\ \text{mm}}{169.0} = \mathbf{0.118\ mm}$$

(a) Velocity zero times.

$$v = \frac{dx}{dt} = 6t^2 - 18t + 12 = 6(t^2 - 3t + 2) = 6(t-1)(t-2)$$

Setting $v = 0$: $\;\mathbf{t = 1\ s\ and\ t = 2\ s}$.

(b) Total distance, $t = 0$ to $3\ \text{s}$. The particle reverses direction at $t = 1$ and $t = 2$, so compute positions at the breakpoints:

$$x(0) = 0$$ $$x(1) = 2 - 9 + 12 = 5\ \text{m}$$ $$x(2) = 16 - 36 + 24 = 4\ \text{m}$$ $$x(3) = 54 - 81 + 36 = 9\ \text{m}$$

Distances per segment:

$$|x(1)-x(0)| = 5\ \text{m}, \quad |x(2)-x(1)| = 1\ \text{m}, \quad |x(3)-x(2)| = 5\ \text{m}$$

Total distance $= 5 + 1 + 5 = \mathbf{11\ m}$.

(a) Principle of linear impulse and momentum. The linear impulse of the net force acting on a particle over a time interval equals the change in its linear momentum:

$$\int_{t_1}^{t_2} \mathbf{F}\,dt = m\mathbf{v}_2 - m\mathbf{v}_1$$

(b) Average braking force. Convert speed: $v_1 = 54\ \text{km/h} = \dfrac{54 \times 1000}{3600} = 15\ \text{m/s}$, and $v_2 = 0$.

For a constant force $F$ over $\Delta t = 4\ \text{s}$:

$$-F\,\Delta t = m(v_2 - v_1)$$ $$F = \frac{m(v_1 - v_2)}{\Delta t} = \frac{1200(15 - 0)}{4} = \frac{18000}{4} = \mathbf{4500\ N\ (4.5\ kN)}$$

Take the common direction of motion as positive. Before impact: $v_A = 6\ \text{m/s}$, $v_B = 2\ \text{m/s}$.

Conservation of linear momentum:

$$m_A v_A + m_B v_B = m_A v_A' + m_B v_B'$$ $$3(6) + 2(2) = 3 v_A' + 2 v_B' \;\Rightarrow\; 18 + 4 = 3 v_A' + 2 v_B'$$ $$3 v_A' + 2 v_B' = 22 \quad (1)$$

Coefficient of restitution:

$$e = \frac{v_B' - v_A'}{v_A - v_B} \;\Rightarrow\; 0.7 = \frac{v_B' - v_A'}{6 - 2}$$ $$v_B' - v_A' = 0.7 \times 4 = 2.8 \quad (2)$$

From (2): $v_B' = v_A' + 2.8$. Substitute into (1):

$$3 v_A' + 2(v_A' + 2.8) = 22 \;\Rightarrow\; 5 v_A' + 5.6 = 22$$ $$5 v_A' = 16.4 \;\Rightarrow\; v_A' = \mathbf{3.28\ m/s}$$ $$v_B' = 3.28 + 2.8 = \mathbf{6.08\ m/s}$$

Both spheres move in the original direction; $B$ is now faster than $A$, so they separate. (Momentum check: $3(3.28) + 2(6.08) = 9.84 + 12.16 = 22$ ✓.)

In $n$-$t$ coordinates the acceleration has two components:

Tangential component (rate of change of speed):

$$a_t = \dot{v} = \mathbf{2\ m/s^2}$$

Normal (centripetal) component:

$$a_n = \frac{v^2}{\rho} = \frac{20^2}{150} = \frac{400}{150} = 2.667\ \text{m/s}^2$$

Magnitude of total acceleration:

$$a = \sqrt{a_t^2 + a_n^2} = \sqrt{2^2 + 2.667^2} = \sqrt{4 + 7.11} = \sqrt{11.11} = \mathbf{3.33\ m/s^2}$$

Direction (angle from the tangential direction toward the centre of the curve):

$$\phi = \tan^{-1}\!\left(\frac{a_n}{a_t}\right) = \tan^{-1}\!\left(\frac{2.667}{2}\right) = \tan^{-1}(1.333) = \mathbf{53.1^\circ\ from\ the\ tangent\ (toward\ the\ centre)}$$

(a) Angular velocity. For rolling without slipping, the contact point is the instantaneous centre of zero velocity (IC), located at the bottom of the wheel a distance $r$ below the centre. The centre velocity is:

$$v_C = \omega r \;\Rightarrow\; \omega = \frac{v_C}{r} = \frac{4}{0.5} = \mathbf{8\ rad/s}$$

(b) Velocity of the topmost point. The top point is a distance $2r$ above the IC. Its speed:

$$v_{\text{top}} = \omega (2r) = 8 \times (2 \times 0.5) = 8 \times 1.0 = \mathbf{8\ m/s}$$

directed horizontally in the direction of motion. (This is twice the velocity of the centre, as expected for a rolling wheel.)

(a) Simple harmonic motion. SHM is a periodic to-and-fro motion in which the restoring force (or acceleration) is directly proportional to the displacement from a fixed equilibrium position and is always directed toward that position. Its governing differential equation is:

$$\ddot{x} + \omega_n^2 x = 0$$

where $\omega_n$ is the natural circular frequency. The general solution is $x(t) = X\sin(\omega_n t + \phi)$.

(b) Simple pendulum. For small oscillations the period is:

$$T = 2\pi\sqrt{\frac{L}{g}} = 2\pi\sqrt{\frac{0.8}{9.81}} = 2\pi\sqrt{0.08155} = 2\pi(0.2856) = \mathbf{1.794\ s}$$

Frequency:

$$f = \frac{1}{T} = \frac{1}{1.794} = \mathbf{0.557\ Hz}$$