BE Civil Engineering (IOE, TU) Applied Mechanics - Dynamics (IOE, CE 452) Question Paper 2080 Nepal

Set up coordinate axes with origin at the launch point (top of building), $x$ horizontal in the direction of throw, $y$ vertically upward.

Initial velocity components:

$$v_{0x} = 25\cos 35^\circ = 25(0.81915) = 20.479\ \text{m/s}$$ $$v_{0y} = 25\sin 35^\circ = 25(0.57358) = 14.339\ \text{m/s}$$

(a) Maximum height above ground

At the peak $v_y = 0$. Rise above launch point:

$$h = \frac{v_{0y}^2}{2g} = \frac{14.339^2}{2(9.81)} = \frac{205.61}{19.62} = 10.48\ \text{m}$$

Height above ground $= 40 + 10.48 = \mathbf{50.48\ m}$.

(b) Total time of flight

Taking the ground as $y = -40\ \text{m}$ relative to the launch point:

$$y = v_{0y}t - \tfrac{1}{2}g t^2 \Rightarrow -40 = 14.339\,t - 4.905\,t^2$$ $$4.905\,t^2 - 14.339\,t - 40 = 0$$ $$t = \frac{14.339 \pm \sqrt{14.339^2 + 4(4.905)(40)}}{2(4.905)} = \frac{14.339 \pm \sqrt{205.61 + 784.8}}{9.81}$$ $$t = \frac{14.339 \pm \sqrt{990.41}}{9.81} = \frac{14.339 \pm 31.471}{9.81}$$

Taking the positive root: $t = \dfrac{45.810}{9.81} = \mathbf{4.67\ s}$.

(c) Horizontal range from foot of building

$$R = v_{0x}\,t = 20.479 \times 4.67 = \mathbf{95.6\ m}$$

(d) Impact velocity

Horizontal: $v_x = v_{0x} = 20.479\ \text{m/s}$.

Vertical: $v_y = v_{0y} - g t = 14.339 - 9.81(4.67) = 14.339 - 45.81 = -31.47\ \text{m/s}$ (downward).

Magnitude:

$$v = \sqrt{20.479^2 + 31.47^2} = \sqrt{419.4 + 990.4} = \sqrt{1409.8} = \mathbf{37.5\ m/s}$$

Direction below horizontal:

$$\theta = \tan^{-1}\!\left(\frac{31.47}{20.479}\right) = \tan^{-1}(1.537) = \mathbf{56.9^\circ\ below\ horizontal}$$

Check the driving tendency (assume B descends, A moves up the incline).

Weight of B: $W_B = 20(9.81) = 196.2\ \text{N}$.

Gravity component of A along incline (down-slope): $W_A\sin 25^\circ = 30(9.81)(0.42262) = 124.4\ \text{N}$.

Since $196.2\ \text{N} > 124.4\ \text{N}$, B descends and A slides up the incline. Friction on A acts down the incline (opposing motion).

Normal force on A: $N = W_A\cos 25^\circ = 30(9.81)(0.90631) = 266.7\ \text{N}$. Friction: $f = \mu_k N = 0.20(266.7) = 53.34\ \text{N}$.

(b) Acceleration — equations of motion (let $a$ be common acceleration, $T$ the tension):

Block B (down positive): $\;m_B g - T = m_B a$

$$196.2 - T = 20a \quad (1)$$

Block A (up-incline positive): $\;T - W_A\sin 25^\circ - f = m_A a$

$$T - 124.4 - 53.34 = 30a \quad (2)$$

Add (1) and (2):

$$196.2 - 124.4 - 53.34 = (20+30)a$$ $$18.46 = 50a \Rightarrow a = \mathbf{0.369\ m/s^2}$$

(c) Tension from (1):

$$T = 196.2 - 20(0.369) = 196.2 - 7.38 = \mathbf{188.8\ N}$$

(d) Speed after descending $1.5\ \text{m}$ (from rest, $v^2 = u^2 + 2as$):

$$v = \sqrt{2(0.369)(1.5)} = \sqrt{1.107} = \mathbf{1.05\ m/s}$$

Work-energy principle: $\;T_1 + \sum U_{1\to 2} = T_2$, where $T = \tfrac{1}{2}mv^2$.

(a) From $x = 0.25\ \text{m}$ (rest) to $x = 0$.

Spring work (spring releases stored energy as collar moves toward natural length, so work is positive):

$$U_{spring} = \tfrac{1}{2}k x_1^2 - \tfrac{1}{2}k x_2^2 = \tfrac{1}{2}(600)(0.25^2) - 0 = \tfrac{1}{2}(600)(0.0625) = 18.75\ \text{J}$$

Friction work over $0.25\ \text{m}$ (negative):

$$U_{fric} = -8(0.25) = -2.0\ \text{J}$$

Energy balance with $T_1 = 0$:

$$0 + 18.75 - 2.0 = \tfrac{1}{2}(4)v^2$$ $$16.75 = 2 v^2 \Rightarrow v^2 = 8.375 \Rightarrow v = \mathbf{2.89\ m/s}$$

(b) Maximum travel $d$ past $x = 0$.

Let the collar momentarily stop at distance $d$ on the far side (spring now stretched by $d$). Apply work-energy from release ($x_1 = 0.25$, rest) to the far stopping point ($x_2 = d$, rest); total path length is $(0.25 + d)$.

Spring: $U_{spring} = \tfrac{1}{2}(600)(0.25)^2 - \tfrac{1}{2}(600)d^2 = 18.75 - 300d^2$

Friction (acts over total distance travelled $0.25 + d$):

$$U_{fric} = -8(0.25 + d) = -2.0 - 8d$$

With $T_1 = T_2 = 0$:

$$18.75 - 300d^2 - 2.0 - 8d = 0$$ $$300d^2 + 8d - 16.75 = 0$$ $$d = \frac{-8 \pm \sqrt{8^2 + 4(300)(16.75)}}{2(300)} = \frac{-8 \pm \sqrt{64 + 20100}}{600} = \frac{-8 \pm 142.0}{600}$$

Positive root: $d = \dfrac{134.0}{600} = \mathbf{0.223\ m}$.

The collar travels about $0.223\ \text{m}$ beyond the natural-length position before stopping (less than $0.25\ \text{m}$, as expected because friction dissipates energy).

Take rightward as positive. Given $v_A = +6\ \text{m/s}$, $v_B = -2\ \text{m/s}$, $m_A = 3\ \text{kg}$, $m_B = 5\ \text{kg}$, $e = 0.7$.

Conservation of linear momentum:

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

Restitution equation:

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

(a) Solve (1) and (2). From (2): $v_B' = v_A' + 5.6$. Substitute into (1):

$$3 v_A' + 5(v_A' + 5.6) = 8 \Rightarrow 8 v_A' + 28 = 8 \Rightarrow 8 v_A' = -20$$ $$v_A' = -2.5\ \text{m/s} \quad(\textbf{2.5 m/s to the left})$$ $$v_B' = -2.5 + 5.6 = +3.1\ \text{m/s} \quad(\textbf{3.1 m/s to the right})$$

(b) Kinetic energy lost.

Before: $T_1 = \tfrac{1}{2}(3)(6^2) + \tfrac{1}{2}(5)(2^2) = 54 + 10 = 64\ \text{J}$.

After: $T_2 = \tfrac{1}{2}(3)(2.5^2) + \tfrac{1}{2}(5)(3.1^2) = \tfrac{1}{2}(3)(6.25) + \tfrac{1}{2}(5)(9.61) = 9.375 + 24.025 = 33.40\ \text{J}$.

Energy lost: $\Delta T = 64 - 33.40 = \mathbf{30.6\ J}$.

(c) Impulse on $A$ from $B$.

Impulse $= \Delta p_A = m_A(v_A' - v_A) = 3(-2.5 - 6) = 3(-8.5) = -25.5\ \text{N·s}$.

Magnitude $= \mathbf{25.5\ N\cdot s}$, directed to the left (opposing $A$'s original motion).

(a) Moment of inertia of a solid disc about its central axis:

$$I = \tfrac{1}{2}m r^2 = \tfrac{1}{2}(50)(0.40)^2 = \tfrac{1}{2}(50)(0.16) = \mathbf{4.0\ kg\cdot m^2}$$

(b) Angular acceleration. Torque applied by the tangential force:

$$\tau = F r = 120(0.40) = 48\ \text{N·m}$$

From $\tau = I\alpha$:

$$\alpha = \frac{\tau}{I} = \frac{48}{4.0} = \mathbf{12\ rad/s^2}$$

(c) Angular velocity and revolutions after $5\ \text{s}$ (from rest):

$$\omega = \alpha t = 12(5) = \mathbf{60\ rad/s}$$ $$\theta = \tfrac{1}{2}\alpha t^2 = \tfrac{1}{2}(12)(5^2) = \tfrac{1}{2}(12)(25) = 150\ \text{rad}$$

Number of revolutions:

$$N = \frac{\theta}{2\pi} = \frac{150}{6.2832} = \mathbf{23.9\ revolutions}$$

(d) Kinetic energy and work check.

Rotational KE:

$$T = \tfrac{1}{2}I\omega^2 = \tfrac{1}{2}(4.0)(60^2) = \tfrac{1}{2}(4.0)(3600) = \mathbf{7200\ J}$$

Work done by the torque:

$$U = \tau\,\theta = 48 \times 150 = 7200\ \text{J}$$

The work done equals the kinetic energy gained ($7200\ \text{J} = 7200\ \text{J}$), verifying the work-energy principle for rotation.

(a) Integrate the acceleration.

$$v(t) = \int (4t - 6)\,dt = 2t^2 - 6t + C_1$$

At $t=0$, $v = -5 \Rightarrow C_1 = -5$. Thus

$$\boxed{v(t) = 2t^2 - 6t - 5\ \text{m/s}}$$ $$x(t) = \int (2t^2 - 6t - 5)\,dt = \tfrac{2}{3}t^3 - 3t^2 - 5t + C_2$$

At $t=0$, $x = 2 \Rightarrow C_2 = 2$. Thus

$$\boxed{x(t) = \tfrac{2}{3}t^3 - 3t^2 - 5t + 2\ \text{m}}$$

(b) At $t = 4\ \text{s}$:

$$v(4) = 2(16) - 6(4) - 5 = 32 - 24 - 5 = \mathbf{3\ m/s}$$ $$x(4) = \tfrac{2}{3}(64) - 3(16) - 5(4) + 2 = 42.667 - 48 - 20 + 2 = \mathbf{-23.33\ m}$$

(c) Particle at rest when $v = 0$:

$$2t^2 - 6t - 5 = 0$$ $$t = \frac{6 \pm \sqrt{36 + 40}}{4} = \frac{6 \pm \sqrt{76}}{4} = \frac{6 \pm 8.718}{4}$$

Physically valid (positive) root:

$$t = \frac{6 + 8.718}{4} = \mathbf{3.68\ s}$$

(The other root $t = -0.68\ \text{s}$ is rejected as it precedes the start of motion.)

Convert speed: $v = 54\ \text{km/h} = \dfrac{54 \times 1000}{3600} = 15\ \text{m/s}$.

(a) Normal (centripetal) acceleration:

$$a_n = \frac{v^2}{r} = \frac{15^2}{80} = \frac{225}{80} = \mathbf{2.8125\ m/s^2} \approx 2.81\ \text{m/s}^2$$

The tangential acceleration is zero since the speed is constant.

(b) Required friction force (provides the centripetal force):

$$F = m a_n = 1200 \times 2.8125 = \mathbf{3375\ N}$$

(c) Minimum coefficient of static friction.

The maximum available friction is $\mu_s N = \mu_s m g$. For no skidding $F \le \mu_s m g$, so the minimum required is

$$\mu_{s,\min} = \frac{F}{m g} = \frac{3375}{1200 \times 9.81} = \frac{3375}{11772} = \mathbf{0.287}$$

Equivalently $\mu_{s,\min} = \dfrac{v^2}{g r} = \dfrac{225}{9.81 \times 80} = \dfrac{225}{784.8} = 0.287$.

For rolling without slipping the contact point is the instantaneous centre of rotation (IC).

(a) Angular velocity and acceleration. With $v_C = r\omega$ and $a_C = r\alpha$:

$$\omega = \frac{v_C}{r} = \frac{4.5}{0.30} = \mathbf{15\ rad/s} \;(\text{clockwise})$$ $$\alpha = \frac{a_C}{r} = \frac{2.0}{0.30} = \mathbf{6.67\ rad/s^2} \;(\text{clockwise})$$

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

$$v_{top} = \omega (2r) = 15(0.60) = \mathbf{9.0\ m/s} \;(\text{to the right})$$

This is twice the centre velocity, as expected for rolling.

(c) Velocity of the front point (level with the centre). Its distance from the IC at the bottom is

$$d = \sqrt{r^2 + r^2} = r\sqrt{2} = 0.30\sqrt{2} = 0.4243\ \text{m}$$

Speed:

$$v = \omega d = 15(0.4243) = \mathbf{6.36\ m/s}$$

The velocity is perpendicular to the line joining the IC to that point, i.e. directed at $45^\circ$ above the horizontal (up-and-forward).

(a) Natural frequency and period.

$$\omega_n = \sqrt{\frac{k}{m}} = \sqrt{\frac{250}{2.5}} = \sqrt{100} = \mathbf{10\ rad/s}$$ $$T = \frac{2\pi}{\omega_n} = \frac{2\pi}{10} = \mathbf{0.628\ s}$$

The natural frequency in Hz: $f = \dfrac{1}{T} = 1.59\ \text{Hz}$.

(b) Equation of motion. For undamped free vibration released from rest at displacement $X = 40\ \text{mm} = 0.040\ \text{m}$, the amplitude is $0.040\ \text{m}$ and the response is a pure cosine (zero initial velocity):

$$\boxed{x(t) = 0.040\cos(10\,t)\ \text{m}}$$

(c) Maximum velocity and acceleration.

$$v_{\max} = \omega_n X = 10(0.040) = \mathbf{0.40\ m/s}$$ $$a_{\max} = \omega_n^2 X = 10^2(0.040) = 100(0.040) = \mathbf{4.0\ m/s^2}$$

Undamped natural frequency:

$$\omega_n = \sqrt{\frac{k}{m}} = \sqrt{\frac{1800}{5}} = \sqrt{360} = 18.974\ \text{rad/s}$$

(a) Critical damping and damping ratio.

$$c_c = 2\sqrt{k m} = 2\sqrt{1800 \times 5} = 2\sqrt{9000} = 2(94.868) = 189.74\ \text{N·s/m}$$ $$\zeta = \frac{c}{c_c} = \frac{60}{189.74} = \mathbf{0.316}$$

Since $\zeta < 1$, the system is underdamped (oscillatory decay).

(b) Damped natural frequency.

$$\omega_d = \omega_n\sqrt{1 - \zeta^2} = 18.974\sqrt{1 - 0.316^2} = 18.974\sqrt{1 - 0.0999}$$ $$= 18.974\sqrt{0.9001} = 18.974(0.9487) = \mathbf{18.00\ rad/s}$$

(c) Logarithmic decrement and amplitude ratio.

$$\delta = \frac{2\pi\zeta}{\sqrt{1-\zeta^2}} = \frac{2\pi(0.316)}{0.9487} = \frac{1.9855}{0.9487} = \mathbf{2.093}$$

Ratio of successive amplitudes:

$$\frac{x_n}{x_{n+1}} = e^{\delta} = e^{2.093} = \mathbf{8.11}$$

Each oscillation cycle reduces the amplitude to about $1/8.11 \approx 12.3\%$ of the previous peak.

Convert angular speed: $\omega_0 = 300\ \text{rpm} = \dfrac{300 \times 2\pi}{60} = 31.416\ \text{rad/s}$.

(a) Moment of inertia and angular momentum.

$$I = \tfrac{1}{2}m r^2 = \tfrac{1}{2}(40)(0.25)^2 = \tfrac{1}{2}(40)(0.0625) = \mathbf{1.25\ kg\cdot m^2}$$ $$H_0 = I\omega_0 = 1.25 \times 31.416 = \mathbf{39.27\ kg\cdot m^2/s}$$

(b) Braking torque (angular impulse-momentum).

$$\tau\,t = I\omega_0 - I\omega_f, \qquad \omega_f = 0$$ $$\tau = \frac{I\omega_0}{t} = \frac{39.27}{8} = \mathbf{4.91\ N\cdot m}$$

(c) Revolutions before stopping.

Angular deceleration: $\alpha = \dfrac{\omega_0}{t} = \dfrac{31.416}{8} = 3.927\ \text{rad/s}^2$.

Angular displacement (using $\omega_f^2 = \omega_0^2 - 2\alpha\theta$ with $\omega_f = 0$):

$$\theta = \frac{\omega_0^2}{2\alpha} = \frac{31.416^2}{2(3.927)} = \frac{987.0}{7.854} = 125.66\ \text{rad}$$

Number of revolutions:

$$N = \frac{\theta}{2\pi} = \frac{125.66}{6.2832} = \mathbf{20.0\ revolutions}$$

(Equivalently, average speed $\tfrac{1}{2}(31.416)\times 8 = 125.66\ \text{rad}$, confirming the result.)