BE Civil Engineering (IOE, TU) Applied Mechanics - Dynamics (IOE, CE 452) Question Paper 2077 Nepal

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Question

1long10 marks

A ball is thrown from the edge of a cliff that is $48\ \text{m}$ above level ground. The ball leaves the thrower's hand with a speed of $25\ \text{m/s}$ at an angle of $36.87^\circ$ above the horizontal (so that $\sin\theta = 0.6$ , $\cos\theta = 0.8$ ). Take $g = 9.81\ \text{m/s}^2$ and neglect air resistance.

(a) Determine the maximum height reached by the ball above the launch point and above the ground.

(b) Find the total time of flight until the ball strikes the ground at the base of the cliff.

(c) Determine the horizontal range measured from the base of the cliff.

(d) Find the magnitude and direction of the velocity of the ball just before it hits the ground.

projectile-motionkinematics-of-particlescurvilinear-motion

Answer 1

Set up the components of the initial velocity.

v_{0x} = v_0\cos\theta = 25(0.8) = 20\ \text{m/s}, \qquad v_{0y} = v_0\sin\theta = 25(0.6) = 15\ \text{m/s}

Take the launch point as origin, $x$ horizontal (positive in direction of throw), $y$ vertical upward. Acceleration: $a_x = 0$ , $a_y = -g = -9.81\ \text{m/s}^2$ .

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

v_y^2 = v_{0y}^2 - 2g\,h_{max} \Rightarrow h_{max} = \frac{v_{0y}^2}{2g} = \frac{15^2}{2(9.81)} = \frac{225}{19.62} = 11.47\ \text{m}

Height above ground $= 48 + 11.47 = \mathbf{59.47\ m}$ (above launch point: $\mathbf{11.47\ m}$ ).

(b) Total time of flight. Ground is at $y = -48\ \text{m}$ :

y = v_{0y}t - \tfrac{1}{2}g t^2 \Rightarrow -48 = 15t - 4.905 t^2

4.905 t^2 - 15 t - 48 = 0

t = \frac{15 \pm \sqrt{15^2 + 4(4.905)(48)}}{2(4.905)} = \frac{15 \pm \sqrt{225 + 941.76}}{9.81} = \frac{15 \pm 34.16}{9.81}

Taking the positive root: $t = \dfrac{49.16}{9.81} = \mathbf{5.01\ s}$ .

(c) Horizontal range from base of cliff.

x = v_{0x}\,t = 20(5.01) = \mathbf{100.2\ m}

(d) Velocity at impact.

v_x = 20\ \text{m/s (constant)}

v_y = v_{0y} - g t = 15 - 9.81(5.01) = 15 - 49.15 = -34.15\ \text{m/s}

Magnitude:

v = \sqrt{20^2 + 34.15^2} = \sqrt{400 + 1166.2} = \sqrt{1566.2} = \mathbf{39.58\ m/s}

Direction below horizontal:

\alpha = \tan^{-1}\!\left(\frac{34.15}{20}\right) = \mathbf{59.65^\circ\ below\ the\ horizontal}

Answer 2

Forces on the incline. Normal force $N = mg\cos 30^\circ$ , friction $f = \mu_k N$ opposing motion.

N = 12(9.81)\cos 30^\circ = 117.72(0.8660) = 101.95\ \text{N}

f = 0.25(101.95) = 25.49\ \text{N}

(a) Work-energy from A to B (block starts from rest, $v_A = 0$ ):

Work by gravity (component along incline): $W_g = mg\sin 30^\circ \cdot d = 117.72(0.5)(6) = 353.16\ \text{J}$ .

Work by friction: $W_f = -f\,d = -25.49(6) = -152.94\ \text{J}$ .

\tfrac{1}{2}m v_B^2 - 0 = W_g + W_f = 353.16 - 152.94 = 200.22\ \text{J}

v_B^2 = \frac{2(200.22)}{12} = 33.37\ \text{m}^2/\text{s}^2 \Rightarrow v_B = \mathbf{5.78\ m/s}

(b) On the horizontal surface. Normal force $N' = mg = 117.72\ \text{N}$ , friction $f' = 0.25(117.72) = 29.43\ \text{N}$ . Only friction does work; block decelerates from $v_B$ to rest over distance $s$ :

0 - \tfrac{1}{2}m v_B^2 = -f' s

s = \frac{\tfrac{1}{2}(12)(33.37)}{29.43} = \frac{200.22}{29.43} = \mathbf{6.80\ m}

Answer 3

Take rightward as positive: $u_A = +6\ \text{m/s}$ , $u_B = -4\ \text{m/s}$ , $m_A = 3$ , $m_B = 2$ .

Conservation of linear momentum:

m_A u_A + m_B u_B = m_A v_A + m_B v_B

3(6) + 2(-4) = 3 v_A + 2 v_B \Rightarrow 18 - 8 = 3 v_A + 2 v_B \Rightarrow 3 v_A + 2 v_B = 10 \quad (1)

Restitution equation:

e = \frac{v_B - v_A}{u_A - u_B} \Rightarrow v_B - v_A = e(u_A - u_B) = 0.7(6-(-4)) = 0.7(10) = 7 \quad (2)

(a) Solve (1) and (2). From (2): $v_B = v_A + 7$ . Substitute into (1):

3 v_A + 2(v_A + 7) = 10 \Rightarrow 5 v_A + 14 = 10 \Rightarrow v_A = -0.8\ \text{m/s}

v_B = -0.8 + 7 = 6.2\ \text{m/s}

So A moves left at $0.8\ \text{m/s}$ and B moves right at $6.2\ \text{m/s}$ .

(b) Kinetic energy loss.

KE_i = \tfrac{1}{2}(3)(6^2) + \tfrac{1}{2}(2)(4^2) = 54 + 16 = 70\ \text{J}

KE_f = \tfrac{1}{2}(3)(0.8^2) + \tfrac{1}{2}(2)(6.2^2) = 0.96 + 38.44 = 39.40\ \text{J}

\Delta KE = 70 - 39.40 = \mathbf{30.60\ J\ lost}

(c) Impulse on B. Using B's change in momentum:

J = m_B(v_B - u_B) = 2(6.2 - (-4)) = 2(10.2) = \mathbf{20.4\ N\cdot s\ (to\ the\ right)}

Answer 4

Convert: $40$ rev $= 40(2\pi) = 251.33\ \text{rad}$ .

(a) Angular acceleration and final angular velocity. Starting from rest, $\theta = \tfrac{1}{2}\alpha t^2$ :

\alpha = \frac{2\theta}{t^2} = \frac{2(251.33)}{8^2} = \frac{502.65}{64} = \mathbf{7.85\ rad/s^2}

\omega = \alpha t = 7.85(8) = \mathbf{62.83\ rad/s}

(Check: $\omega = 2\pi(2 \times 40/8) =$ consistent.)

(b) Acceleration of a rim point ( $r = 0.5\ \text{m}$ ).

a_t = r\alpha = 0.5(7.85) = 3.93\ \text{m/s}^2

a_n = r\omega^2 = 0.5(62.83)^2 = 0.5(3947.6) = 1973.8\ \text{m/s}^2

a = \sqrt{a_t^2 + a_n^2} = \sqrt{3.93^2 + 1973.8^2} = \sqrt{15.4 + 3.8958\times10^6} \approx \mathbf{1973.8\ m/s^2}

(The normal component dominates overwhelmingly.)

(c) Deceleration phase. Initial $\omega = 62.83\ \text{rad/s}$ , final $0$ , over $20$ rev $= 125.66\ \text{rad}$ :

\omega_f^2 = \omega^2 + 2\alpha' \theta' \Rightarrow 0 = 62.83^2 + 2\alpha'(125.66)

\alpha' = \frac{-3947.6}{251.33} = -15.71\ \text{rad/s}^2

The magnitude of the angular deceleration is $\mathbf{15.71\ rad/s^2}$ .

Answer 5

(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} = \mathbf{3.18\ Hz}\quad(\omega_n = \mathbf{20\ rad/s})

(b) Damping ratio. Critical damping coefficient:

c_c = 2\sqrt{km} = 2\sqrt{1600(4)} = 2\sqrt{6400} = 2(80) = 160\ \text{N}\cdot\text{s/m}

\zeta = \frac{c}{c_c} = \frac{40}{160} = \mathbf{0.25}

Since $\zeta < 1$ , the system is under-damped.

(c) Damped natural frequency and period.

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

T_d = \frac{2\pi}{\omega_d} = \frac{6.2832}{19.36} = \mathbf{0.3245\ s}

(d) Logarithmic decrement.

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

Ratio of successive amplitudes:

\frac{x_n}{x_{n+1}} = e^{\delta} = e^{1.622} = \mathbf{5.06}

Answer 6

(a) Differentiate.

v(t) = \frac{dx}{dt} = 6t^2 - 18t + 12\ \text{m/s}, \qquad a(t) = \frac{dv}{dt} = 12t - 18\ \text{m/s}^2

(b) At rest when $v = 0$ :

6t^2 - 18t + 12 = 0 \Rightarrow t^2 - 3t + 2 = 0 \Rightarrow (t-1)(t-2) = 0

So $t = \mathbf{1\ s}$ and $t = \mathbf{2\ s}$ .

(c) Total distance, $t=0$ to $t=3$ . The particle reverses at $t=1$ and $t=2$ , so evaluate positions:

x(0) = 5\ \text{m}

x(1) = 2 - 9 + 12 + 5 = 10\ \text{m}

x(2) = 16 - 36 + 24 + 5 = 9\ \text{m}

x(3) = 54 - 81 + 36 + 5 = 14\ \text{m}

Distances over each monotonic segment:

|x(1)-x(0)| = |10-5| = 5\ \text{m}

|x(2)-x(1)| = |9-10| = 1\ \text{m}

|x(3)-x(2)| = |14-9| = 5\ \text{m}

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

Answer 7

Free-body equations. Let $a$ be the common acceleration, $T$ the cord tension.

Block A (hanging, moving down): $m_A g - T = m_A a$ .

Block B (on smooth table, moving horizontally): $T = m_B a$ .

(a) Solve for acceleration. Add the equations:

m_A g = (m_A + m_B)a \Rightarrow a = \frac{m_A g}{m_A + m_B} = \frac{10(9.81)}{10+4} = \frac{98.1}{14} = \mathbf{7.01\ m/s^2}

(b) Tension.

T = m_B a = 4(7.01) = \mathbf{28.03\ N}

(Check with A: $m_A g - T = 98.1 - 28.03 = 70.07 = 10(7.01)$ ✓)

(c) Speed after descending $1.5\ \text{m}$ (from rest, constant $a$ ):

v^2 = u^2 + 2 a s = 0 + 2(7.01)(1.5) = 21.03 \Rightarrow v = \mathbf{4.59\ m/s}

Answer 8

(a) Mass moment of inertia of a solid cylinder about its central axis, $I = \tfrac{1}{2}mR^2$ :

I = \tfrac{1}{2}(20)(0.3)^2 = \tfrac{1}{2}(20)(0.09) = \mathbf{0.9\ kg\cdot m^2}

(b) Angular acceleration. Applied torque $\tau = F R = 50(0.3) = 15\ \text{N}\cdot\text{m}$ .

\alpha = \frac{\tau}{I} = \frac{15}{0.9} = \mathbf{16.67\ rad/s^2}

(c) After $t = 4\ \text{s}$ from rest.

\omega = \alpha t = 16.67(4) = \mathbf{66.67\ rad/s}

\theta = \tfrac{1}{2}\alpha t^2 = \tfrac{1}{2}(16.67)(4^2) = \tfrac{1}{2}(16.67)(16) = 133.3\ \text{rad}

N = \frac{\theta}{2\pi} = \frac{133.3}{6.2832} = \mathbf{21.2\ revolutions}

Answer 9

(a) Components in the normal-tangential (n-t) frame.

Tangential (rate of change of speed):

a_t = \frac{dv}{dt} = \mathbf{3\ m/s^2}

Normal (centripetal):

a_n = \frac{v^2}{\rho} = \frac{25^2}{200} = \frac{625}{200} = \mathbf{3.125\ m/s^2}

(b) Total acceleration.

a = \sqrt{a_t^2 + a_n^2} = \sqrt{3^2 + 3.125^2} = \sqrt{9 + 9.766} = \sqrt{18.766} = \mathbf{4.33\ m/s^2}

Answer 10

Constant acceleration:

a = \frac{v - u}{t} = \frac{20 - 0}{10} = 2\ \text{m/s}^2

(a) Tractive force.

F = ma = 1200(2) = \mathbf{2400\ N}

(b) Power.

Average power $=$ work done / time. Work $= \tfrac{1}{2}mv^2 = \tfrac{1}{2}(1200)(20^2) = 240000\ \text{J}$ .

P_{avg} = \frac{240000}{10} = 24000\ \text{W} = \mathbf{24\ kW}

Maximum power occurs at maximum speed ( $v = 20\ \text{m/s}$ ):

P_{max} = F v = 2400(20) = 48000\ \text{W} = \mathbf{48\ kW}

Answer 11

(a) Assumptions and derivation. A simple pendulum is a point mass (bob) on a light, inextensible string fixed at the top, swinging in a vertical plane with no air resistance, through a small angle. For angular displacement $\theta$ , the restoring tangential force is $-mg\sin\theta$ . Newton's second law along the arc (arc length $s = L\theta$ ):

m L\ddot\theta = -mg\sin\theta

For small $\theta$ , $\sin\theta \approx \theta$ :

\ddot\theta + \frac{g}{L}\theta = 0

This is SHM with $\omega_n = \sqrt{g/L}$ , hence period

T = 2\pi\sqrt{\frac{L}{g}}

(b) Numerical values for $L = 0.8\ \text{m}$ .

T = 2\pi\sqrt{\frac{0.8}{9.81}} = 2\pi\sqrt{0.08155} = 2\pi(0.28557) = \mathbf{1.794\ s}

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

BE Civil Engineering (IOE, TU) Applied Mechanics - Dynamics (IOE, CE 452) Question Paper 2077 Nepal

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