BE Civil Engineering (IOE, TU) Applied Mechanics - Dynamics (IOE, CE 452) Question Paper 2079 Nepal

Set up the components of the initial velocity.

$$v_{0x} = v_0\cos\theta = 25\cos 35^\circ = 25(0.8192) = 20.48\text{ m/s}$$ $$v_{0y} = v_0\sin\theta = 25\sin 35^\circ = 25(0.5736) = 14.34\text{ m/s}$$

Take the launch point as origin, $y$ positive upward, $a_y = -g = -9.81\text{ m/s}^2$.

(a) Maximum height.

At the apex $v_y = 0$:

$$y_{\max,\text{above launch}} = \frac{v_{0y}^2}{2g} = \frac{(14.34)^2}{2(9.81)} = \frac{205.6}{19.62} = 10.48\text{ m}$$

Height above the ground:

$$H = h + 10.48 = 30 + 10.48 = \mathbf{40.48\text{ m}}$$

(b) Total time of flight.

The ball lands at $y = -30\text{ m}$ relative to the launch point:

$$-30 = v_{0y}t - \tfrac{1}{2}g t^2 = 14.34\,t - 4.905\,t^2$$ $$4.905\,t^2 - 14.34\,t - 30 = 0$$

Solve the quadratic:

$$t = \frac{14.34 \pm \sqrt{(14.34)^2 + 4(4.905)(30)}}{2(4.905)} = \frac{14.34 \pm \sqrt{205.6 + 588.6}}{9.81} = \frac{14.34 \pm 28.18}{9.81}$$

Taking the positive root:

$$t = \frac{14.34 + 28.18}{9.81} = \frac{42.52}{9.81} = \mathbf{4.335\text{ s}}$$

(c) Range and impact speed.

Horizontal range:

$$R = v_{0x}\,t = 20.48 \times 4.335 = \mathbf{88.78\text{ m}}$$

Velocity components at impact:

$$v_x = 20.48\text{ m/s (constant)}$$ $$v_y = v_{0y} - g t = 14.34 - 9.81(4.335) = 14.34 - 42.53 = -28.19\text{ m/s}$$

Impact speed:

$$v = \sqrt{v_x^2 + v_y^2} = \sqrt{(20.48)^2 + (28.19)^2} = \sqrt{419.4 + 794.7} = \sqrt{1214.1} = \mathbf{34.84\text{ m/s}}$$

Direction below horizontal: $\alpha = \tan^{-1}(28.19/20.48) = 54.0^\circ$ below the horizontal.

(a) Free-body diagrams.

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

   N
   |        Block A
   o----> T
   |  f <--
   |
   v W_A

Block B (hanging): weight $W_B = m_B g$ down, tension $T$ up.

   | T (up)
   o  Block B
   |
   v W_B (down)

Block B is heavier in the driving sense, so A moves toward the pulley and B moves down with common acceleration $a$.

(b) Equations of motion.

Normal force on A: $N = m_A g = 12(9.81) = 117.72\text{ N}$.

Friction on A: $f = \mu_k N = 0.25(117.72) = 29.43\text{ N}$.

Block A (horizontal):

$$T - f = m_A a \;\Rightarrow\; T - 29.43 = 12a \quad (1)$$

Block B (vertical, down positive):

$$m_B g - T = m_B a \;\Rightarrow\; 78.48 - T = 8a \quad (2)$$

Add (1) and (2):

$$78.48 - 29.43 = (12 + 8)a$$ $$49.05 = 20a \;\Rightarrow\; a = \mathbf{2.453\text{ m/s}^2}$$

(c) Tension.

From (2): $T = m_B g - m_B a = 78.48 - 8(2.453) = 78.48 - 19.62 = \mathbf{58.86\text{ N}}$

Check with (1): $T = 29.43 + 12(2.453) = 29.43 + 29.44 = 58.87\text{ N}$ (consistent).

Work-energy principle: $U_{1\to2} = \Delta KE = \tfrac12 m v_2^2 - \tfrac12 m v_1^2$, with $v_1 = 0$.

The collar moves from $x_1 = 0.30\text{ m}$ to $x_2 = 0$, a displacement of $0.30\text{ m}$ in the direction of decreasing $x$ (same direction as $P$).

Work done by force $P$ (acts in direction of motion):

$$U_P = P\,d = 60(0.30) = 18.0\text{ J}$$

Work done by the spring (spring is releasing stored energy as it returns toward unstretched length, so it does positive work on the collar):

$$U_{spring} = \tfrac12 k x_1^2 - \tfrac12 k x_2^2 = \tfrac12(800)(0.30)^2 - 0 = \tfrac12(800)(0.09) = 36.0\text{ J}$$

The rod is smooth, so friction work is zero. Total work:

$$U_{1\to2} = 18.0 + 36.0 = 54.0\text{ J}$$

Apply the principle:

$$54.0 = \tfrac12 (5) v_2^2$$ $$v_2^2 = \frac{2(54.0)}{5} = \frac{108}{5} = 21.6\text{ m}^2/\text{s}^2$$ $$v_2 = \sqrt{21.6} = \mathbf{4.648\text{ m/s}}$$

Given: $u_1 = +3.0\text{ m/s}$, $u_2 = -2.0\text{ m/s}$, $m_1 = 0.20\text{ kg}$, $m_2 = 0.30\text{ kg}$, $e = 0.80$.

(a) Conservation of linear momentum:

$$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$$ $$0.20(3.0) + 0.30(-2.0) = 0.20 v_1 + 0.30 v_2$$ $$0.60 - 0.60 = 0.20 v_1 + 0.30 v_2$$ $$0 = 0.20 v_1 + 0.30 v_2 \quad (1)$$

Restitution equation:

$$e = \frac{v_2 - v_1}{u_1 - u_2} \;\Rightarrow\; v_2 - v_1 = e(u_1 - u_2) = 0.80\,(3.0 - (-2.0)) = 0.80(5.0) = 4.0$$ $$v_2 - v_1 = 4.0 \quad (2)$$

From (1): $0.20 v_1 = -0.30 v_2 \Rightarrow v_1 = -1.5 v_2$.

Substitute into (2): $v_2 - (-1.5 v_2) = 4.0 \Rightarrow 2.5 v_2 = 4.0 \Rightarrow v_2 = 1.6\text{ m/s}$.

Then $v_1 = -1.5(1.6) = -2.4\text{ m/s}$.

$$\boxed{v_1 = -2.40\text{ m/s (to the left)}, \quad v_2 = +1.60\text{ m/s (to the right)}}$$

The balls have reversed direction, which is physically consistent.

(b) Kinetic energy lost.

Before:

$$KE_i = \tfrac12(0.20)(3.0)^2 + \tfrac12(0.30)(2.0)^2 = 0.90 + 0.60 = 1.50\text{ J}$$

After:

$$KE_f = \tfrac12(0.20)(2.4)^2 + \tfrac12(0.30)(1.6)^2 = \tfrac12(0.20)(5.76) + \tfrac12(0.30)(2.56)$$ $$KE_f = 0.576 + 0.384 = 0.960\text{ J}$$

Energy lost:

$$\Delta KE = 1.50 - 0.960 = \mathbf{0.540\text{ J}}$$

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

$$I = \tfrac12 M r^2 = \tfrac12 (40)(0.50)^2 = \tfrac12(40)(0.25) = \mathbf{5.0\text{ kg·m}^2}$$

(b) Angular acceleration from the rotational equation of motion $\sum T = I\alpha$.

Driving torque from the belt force: $T_F = F r = 60(0.50) = 30.0\text{ N·m}$.

Net torque: $T_{net} = T_F - T_{friction} = 30.0 - 4.0 = 26.0\text{ N·m}$.

$$\alpha = \frac{T_{net}}{I} = \frac{26.0}{5.0} = \mathbf{5.20\text{ rad/s}^2}$$

(c) Angular velocity and revolutions after $t = 6.0\text{ s}$ (constant $\alpha$, $\omega_0 = 0$).

Angular velocity:

$$\omega = \omega_0 + \alpha t = 0 + 5.20(6.0) = \mathbf{31.2\text{ rad/s}}$$

Angular displacement:

$$\theta = \omega_0 t + \tfrac12 \alpha t^2 = 0 + \tfrac12(5.20)(6.0)^2 = \tfrac12(5.20)(36) = 93.6\text{ rad}$$

Number of revolutions:

$$N = \frac{\theta}{2\pi} = \frac{93.6}{6.2832} = \mathbf{14.9\text{ revolutions}}$$

(a) Integrate the acceleration.

Velocity ($v = v_0 + \int_0^t a\,dt$):

$$v = 5 + \int_0^t (6t - 4)\,dt = 5 + \left[3t^2 - 4t\right]_0^t = \mathbf{3t^2 - 4t + 5\ \text{m/s}}$$

Position ($s = s_0 + \int_0^t v\,dt$):

$$s = 2 + \int_0^t (3t^2 - 4t + 5)\,dt = 2 + \left[t^3 - 2t^2 + 5t\right]_0^t = \mathbf{t^3 - 2t^2 + 5t + 2\ \text{m}}$$

(b) Evaluate at $t = 3\text{ s}$.

Velocity:

$$v(3) = 3(3)^2 - 4(3) + 5 = 27 - 12 + 5 = \mathbf{20\text{ m/s}}$$

Position:

$$s(3) = (3)^3 - 2(3)^2 + 5(3) + 2 = 27 - 18 + 15 + 2 = \mathbf{26\text{ m}}$$

(a) Natural frequency.

Circular (angular) natural frequency:

$$\omega_n = \sqrt{\frac{k}{m}} = \sqrt{\frac{1600}{4}} = \sqrt{400} = \mathbf{20.0\text{ rad/s}}$$

Natural frequency:

$$f_n = \frac{\omega_n}{2\pi} = \frac{20.0}{6.2832} = \mathbf{3.183\text{ Hz}}$$

Period:

$$T = \frac{1}{f_n} = \frac{2\pi}{\omega_n} = \frac{6.2832}{20.0} = \mathbf{0.3142\text{ s}}$$

(b) Equation of motion.

Released from rest at amplitude $A = 0.05\text{ m}$, so:

$$\boxed{x(t) = 0.05\cos(20.0\,t)\ \text{m}}$$

Maximum velocity:

$$v_{\max} = A\,\omega_n = 0.05(20.0) = \mathbf{1.0\text{ m/s}}$$

Maximum acceleration:

$$a_{\max} = A\,\omega_n^2 = 0.05(20.0)^2 = 0.05(400) = \mathbf{20.0\text{ m/s}^2}$$

(a) Normal (centripetal) acceleration.

Speed is constant, so the tangential acceleration is zero. Only normal acceleration acts:

$$a_n = \frac{v^2}{R} = \frac{(20)^2}{80} = \frac{400}{80} = \mathbf{5.0\text{ m/s}^2}$$

(b) Required friction force. This supplies the centripetal force (Newton's second law in the normal direction):

$$F_f = m\,a_n = 1200(5.0) = \mathbf{6000\text{ N}}$$

(c) Minimum coefficient of static friction.

On a flat road the normal reaction equals the weight: $N = mg = 1200(9.81) = 11772\text{ N}$.

For no skidding: $F_f \le \mu_s N$, so the minimum is

$$\mu_{s,\min} = \frac{F_f}{N} = \frac{6000}{11772} = 0.5097$$

Equivalently $\mu_{s,\min} = \dfrac{v^2}{gR} = \dfrac{400}{9.81(80)} = \dfrac{400}{784.8} = \mathbf{0.510}$

(a) Rolling-without-slipping relations.

For pure rolling, the contact point is the instantaneous centre of zero velocity, so:

$$v_C = \omega r \;\Rightarrow\; \omega = \frac{v_C}{r} = \frac{6.0}{0.40} = \mathbf{15.0\text{ rad/s}}$$ $$a_C = \alpha r \;\Rightarrow\; \alpha = \frac{a_C}{r} = \frac{2.0}{0.40} = \mathbf{5.0\text{ rad/s}^2}$$

(b) Velocity of the topmost point.

The top point is a distance $2r$ above the instantaneous centre (the contact point), so its speed is:

$$v_{top} = \omega(2r) = 15.0(2 \times 0.40) = 15.0(0.80) = \mathbf{12.0\text{ m/s}}$$

Equivalently $v_{top} = v_C + \omega r = 6.0 + 6.0 = 12.0\text{ m/s}$ (directed horizontally in the direction of motion).

(c) Velocity of the contact point.

$$v_{contact} = \mathbf{0\text{ m/s}}$$

Because the wheel rolls without slipping, the point of the wheel touching the ground is instantaneously at rest relative to the ground; it is the instantaneous centre of rotation. (Note: it still has a non-zero acceleration directed toward the centre, equal to $\omega^2 r$.)

Work-energy principle for a rigid body: $U_{1\to2} = \Delta KE$, with the cylinder starting from rest.

Work done by gravity (drop in height $= s\sin\theta$):

$$U = mg\,s\sin\theta = 10(9.81)(4.0)\sin 25^\circ = 392.4 \times 0.4226 = 165.8\text{ J}$$

Friction does no work in rolling without slipping (contact point has zero velocity).

Total kinetic energy of a rolling body (translation + rotation):

$$KE = \tfrac12 m v^2 + \tfrac12 I\omega^2$$

For a solid cylinder $I = \tfrac12 m r^2$, and rolling gives $\omega = v/r$:

$$KE = \tfrac12 m v^2 + \tfrac12\left(\tfrac12 m r^2\right)\frac{v^2}{r^2} = \tfrac12 m v^2 + \tfrac14 m v^2 = \tfrac34 m v^2$$

Apply the principle:

$$165.8 = \tfrac34 (10) v^2 = 7.5\,v^2$$ $$v^2 = \frac{165.8}{7.5} = 22.11\text{ m}^2/\text{s}^2$$ $$v = \sqrt{22.11} = \mathbf{4.702\text{ m/s}}$$

(As a check, $v = \sqrt{\tfrac{4}{3}g s\sin\theta} = \sqrt{1.3333 \times 9.81 \times 4.0 \times 0.4226} = \sqrt{22.11} = 4.70\text{ m/s}$.)

(a) Critical damping and damping ratio.

Undamped natural frequency:

$$\omega_n = \sqrt{\frac{k}{m}} = \sqrt{\frac{2000}{5}} = \sqrt{400} = 20.0\text{ rad/s}$$

Critical damping coefficient:

$$c_c = 2\sqrt{km} = 2\sqrt{2000 \times 5} = 2\sqrt{10000} = 2(100) = \mathbf{200\text{ N·s/m}}$$

Damping ratio:

$$\zeta = \frac{c}{c_c} = \frac{30}{200} = \mathbf{0.15}$$

Since $\zeta < 1$, the system is underdamped (it oscillates with decaying amplitude).

(b) Damped natural frequency.

$$\omega_d = \omega_n\sqrt{1 - \zeta^2} = 20.0\sqrt{1 - (0.15)^2} = 20.0\sqrt{1 - 0.0225} = 20.0\sqrt{0.9775}$$ $$\omega_d = 20.0(0.9887) = \mathbf{19.77\text{ rad/s}}$$

(c) Logarithmic decrement.

$$\delta = \frac{2\pi\zeta}{\sqrt{1 - \zeta^2}} = \frac{2\pi(0.15)}{0.9887} = \frac{0.9425}{0.9887} = \mathbf{0.9532}$$