1. Problem 1 State the problem: A flywheel of weight 2006.55 lb has a radius of gyration 15.24 cm; the shaft journal rim diameter is 3.048 cm and coefficient of friction 0.06. The flywheel slows from 2 rps to 1 rps after driving force is withdrawn. Find: (a) Energy loss of the wheel. (b) Moment opposing motion. Step 1: Convert units and find mass moment of inertia $I$ Weight $W = 2006.55$ lb $\Rightarrow$ mass $m = \frac{W}{g} = \frac{2006.55}{32.174} = 62.38$ slugs (approx). Radius of gyration $k = 15.24$ cm = 0.1524 m. Moment of inertia $I = m k^2 = 62.38 \times (0.1524)^2 = 1.447$ slug$\,m^2$ (converted units for consistency). Step 2: Calculate initial and final angular velocities ($\omega_1$ and $\omega_2$) Given in revolutions per second (rps): 2 rps and 1 rps. Angular velocity $\omega = 2 \pi \times \text{rps}$ $\omega_1 = 2 \pi \times 2 = 4\pi$ rad/s $\omega_2 = 2 \pi \times 1 = 2\pi$ rad/s Step 3: Calculate the energy loss Kinetic energy initial $KE_1 = \frac{1}{2} I \omega_1^2 = \frac{1}{2} \times 1.447 \times (4\pi)^2$ $= 0.7235 \times 16 \pi^2 = 0.7235 \times 157.91 = 114.2$ (slug m$^2$/s$^2$) Kinetic energy final $KE_2 = \frac{1}{2} I \omega_2^2 = 0.7235 \times (2\pi)^2 = 0.7235 \times 39.48 = 28.56$ Energy loss $\Delta E = KE_1 - KE_2 = 114.2 - 28.56 = 85.64$ (slug m$^2$/s$^2$), equivalently joules. Step 4: Calculate moment opposing motion The friction force $F_f = \mu N$, where $N$ is normal force (weight component on friction area). Journal rim diameter $d = 3.048$ cm = 0.03048 m, radius $r = 0.01524$ m. Opposing torque $T = F_f \times r$, and the resisting moment equals the torque that causes deceleration from 2 rps to 1 rps. Angular acceleration $\alpha = \frac{\Delta \omega}{\Delta t}$, but time $\Delta t$ not given; alternatively, use energy loss relation to find average resisting moment. Average resisting torque $T = \frac{\Delta E}{\Delta \theta}$. Angular displacement $\Delta \theta$ for slowing from 2 rps to 1 rps depends on unknown time, so assume constant friction torque equal to $T = I \alpha$. Without time or angular deceleration, exact opposing moment cannot be determined numerically here without extra assumptions. --- 2. Problem 2 State the problem: Mechanical press punches 6 holes/min on 0.25 cm thick plate; hole diameter 0.25 cm; ultimate shear strength 60932.64 psi; operating speed slows from 200 rpm to 180 rpm; flywheel mean diameter 1 m; rim width 3x rim thickness; hub and arm mass 5% of rim mass at mean diameter; cast-iron density 7200 kg/m$^3$. Find: (a) Power in kW required. (b) Width and thickness of flywheel rim in mm. Step 1: Calculate shear area and force per hole punched Hole diameter $d = 0.25$ cm = 0.0025 m Plate thickness $t = 0.25$ cm = 0.0025 m Shear area $A = \pi d t = \pi \times 0.0025 \times 0.0025 = 1.9635 \times 10^{-5}$ m$^2$ Step 2: Convert ultimate strength from psi to Pa $1$ psi = 6894.76 Pa Ultimate shear strength $\tau_u = 60932.64 \times 6894.76 = 4.20 \times 10^8$ Pa Step 3: Calculate force needed to punch one hole Force $F = \tau_u \times A = 4.20 \times 10^8 \times 1.9635 \times 10^{-5} = 8247$ N Step 4: Calculate work done per hole Assume punching displacement equals plate thickness $t = 0.0025$ m Work per hole $W = F \times t = 8247 \times 0.0025 = 20.62$ J Step 5: Calculate power needed for 6 holes per minute Holes per minute = 6 Power $P = \frac{6 \times 20.62}{60} = 2.06$ W Step 6: Consider speed change from 200 to 180 rpm and adjust power accordingly Mechanical press slows down by 10%, stress work remains approximately constant. Estimate power required is around $P \approx 2.06$ W or 0.00206 kW Step 7: Calculate dimensions of flywheel rim Assume rim volume $V = \pi d w t$ where $w$ is width and $t$ is thickness, with $w = 3 t$ Mass $m = \rho V$, density $\rho = 7200$ kg/m$^3$ Litited data provided for exact dimensions; more info needed for full calculation. --- 3. Problem 3 State the problem: Differential band brake with angle of contact $\theta = 380^\circ$, torque $T=7000$ in-lbs (791000 N mm), band bearing on 14 in (355.6 mm) diameter drum; friction coefficient 0.3; dimensions given. Find: horsepower and operating force $F$. Step 1: Convert torque to Nm $T = 7000$ in-lbs $1$ in-lb = 0.113 Nm $T = 7000 \times 0.113 = 791$ Nm Step 2: Calculate the friction factor and force using band brake equation $\frac{T}{r} = F (e^{\mu \theta} - 1)$ where $r$ is drum radius, Angle $\theta$ in radians: $380^\circ = \frac{380 \pi}{180} = 6.64$ rad, $\mu = 0.3$, $r = \frac{355.6}{2} = 177.8$ mm = 0.1778 m $F = \frac{T}{r (e^{\mu \theta} - 1)} = \frac{791}{0.1778 (e^{0.3*6.64} - 1)}$ Calculate exponent: $e^{1.992} = 7.34$ So denominator $= 0.1778 \times (7.34 - 1) = 1.144$ Then $F = \frac{791}{1.144} = 691.5$ N Step 3: Calculate power output Power $P = T \times \omega$, assuming operating speed needed (rpm not given), thus horsepower not exactly computable without speed. --- 4. Problem 4 State the problem: Block brake with spring force $A=3034$ N on each lever arm, wheel diameter 356 mm, friction coefficient 0.40, angle of contact 110°. Find energy absorbed by brake. Step 1: Calculate normal force on brake lining (two arms) Total normal force $N = 2 \times A = 2 \times 3034 = 6068$ N Step 2: Calculate friction force $F_f = \mu N = 0.40 \times 6068 = 2427$ N Step 3: Calculate braking torque Radius $r = \frac{356}{2} = 178$ mm = 0.178 m Torque $T = F_f \times r = 2427 \times 0.178 = 431.9$ Nm Step 4: Calculate energy absorbed Assuming brake holds for a certain angular displacement $\theta$, energy $E = T \times \theta$ Since $\theta$ not provided, assume full stop, the exact energy absorbed cannot be calculated without angular displacement. Final answers: Problem 1: (a) Energy loss $\approx 85.64$ J (b) Opposing moment depends on angular deceleration, further data needed. Problem 2: (a) Power $\approx 0.00206$ kW (b) Width and thickness indeterminate without more data. Problem 3: Operating force $F = 691.5$ N Horsepower not computable without speed. Problem 4: Torque $= 431.9$ Nm Energy absorbed dependent on unknown angular displacement.