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A clutch connects shaft A to shaft B. Shaft A rotates at 250 rpm. Shaft B starts from rest and reaches full speed in 8 seconds. Calculate the torque needed to accelerate shaft B.

Physics

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Step 1: Convert given values to SI units.
The external diameter is Do=220 mmD_o = 220 \text{ mm}, so the external radius is Ro=2202=110 mm=0.110 mR_o = \frac{220}{2} = 110 \text{ mm} = 0.110 \text{ m}.
The internal diameter is Di=110 mmD_i = 110 \text{ mm}, so the internal radius is Ri=1102=55 mm=0.055 mR_i = \frac{110}{2} = 55 \text{ mm} = 0.055 \text{ m}.
The speed of shaft A is NA=250 rpmN_A = 250 \text{ rpm}.
The time to reach full speed is t=8 st = 8 \text{ s}.
The coefficient of friction is μ=0.3\mu = 0.3.
The moment of inertia of shaft B is IB=30 kg⋅m2I_B = 30 \text{ kg} \cdot \text{m}^2.

Step 2: Calculate the angular velocity of shaft A.
The angular velocity ωA\omega_A is given by:

ωA=2πNA60\omega_A = \frac{2 \pi N_A}{60} ωA=2π(250)60=500π60=25π3 rad/s\omega_A = \frac{2 \pi (250)}{60} = \frac{500 \pi}{60} = \frac{25 \pi}{3} \text{ rad/s} ωA≈26.1799 rad/s\omega_A \approx 26.1799 \text{ rad/s}

Step 3: Calculate the angular acceleration of shaft B.
Shaft B starts from rest (ωB,initial=0\omega_{B,initial} = 0) and reaches the speed of shaft A (ωB,final=ωA\omega_{B,final} = \omega_A) in t=8t = 8 seconds.

αB=ωB,final−ωB,initialt\alpha_B = \frac{\omega_{B,final} - \omega_{B,initial}}{t} αB=25π3−08=25π24 rad/s2\alpha_B = \frac{\frac{25 \pi}{3} - 0}{8} = \frac{25 \pi}{24} \text{ rad/s}^2 αB≈3.2725 rad/s2\alpha_B \approx 3.2725 \text{ rad/s}^2

Step 4: Calculate the torque transmitted by the clutch.
The torque TT required to accelerate shaft B is given by:

T=IBαBT = I_B \alpha_B T=(30 kg⋅m2)(25π24 rad/s2)T = (30 \text{ kg} \cdot \text{m}^2) \left(\frac{25 \pi}{24} \text{ rad/s}^2\right) T=30×25π24=5×25π4=125π4 N⋅mT = \frac{30 \times 25 \pi}{24} = \frac{5 \times 25 \pi}{4} = \frac{125 \pi}{4} \text{ N} \cdot \text{m} T≈98.1748 N⋅mT \approx 98.1748 \text{ N} \cdot \text{m}

Step 5: Calculate the mean radius for uniform pressure.
For uniform pressure, the mean radius RmR_m is given by:

Rm=23Ro3−Ri3Ro2−Ri2R_m = \frac{2}{3} \frac{R_o^3 - R_i^3}{R_o^2 - R_i^2} Rm=23(0.110 m)3−(0.055 m)3(0.110 m)2−(0.055 m)2R_m = \frac{2}{3} \frac{(0.110 \text{ m})^3 - (0.055 \text{ m})^3}{(0.110 \text{ m})^2 - (0.055 \text{ m})^2} Rm=230.001331−0.0001663750.0121−0.003025R_m = \frac{2}{3} \frac{0.001331 - 0.000166375}{0.0121 - 0.003025}
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