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Andrews [41]
3 years ago
5

The shaft of radius c is subjected to a distributed torque t, measured as torque/unit length of shaft. Shaft A B of length L, fi

xed at B, with x measured from B. Torque t = t sub 0 [1 + (x over L) squared], from t sub 0 at B to 2 t sub 0 at A. The shear modulus is G. Part A Determine the angle of twist at end A. Express your answer as an expression in terms of the variables t0, L, c, and G and any necessary constants.

Physics
1 answer:
tresset_1 [31]3 years ago
7 0

Answer:

Angle of twist;Φ = (to•L²)/Gπc⁴

Explanation:

I have attached the explanation.

However, I used D for radius of shaft against c used in the question. So, if we replace the D with c, we'll have;

Φ = (to•L²)/Gπc⁴

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\begin{gathered} \sigma=\frac{m}{v} \\ m=\sigma\cdot v \\ \text{mass}=(2\cdot10^3\frac{\operatorname{kg}}{m^3})\cdot(0.01m)^3 \\ \text{mass}=(2\cdot10^3\frac{\operatorname{kg}}{m^3})\cdot(1\cdot10^{-6}) \\ \text{mass}=2\cdot10^{-3}\operatorname{kg} \\ \text{mass}=0.002\text{ kg } \\ \text{Area}=(0.01\text{ m}\cdot0.01m)=0.0001m^2 \end{gathered}

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\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ v=\sqrt[]{\frac{2(0.002kg)(9.81\text{ }\frac{m}{s^2})}{(2\cdot10^3\frac{\operatorname{kg}}{m^3})(0.0001m^2)0.8}} \\ v=\sqrt[]{\frac{0.03924\frac{\operatorname{kg}m}{s^2}}{0.16\frac{\operatorname{kg}}{m^{}}}} \\ v=\sqrt[]{0.2452\frac{m^2}{s^2}} \\ v=0.495\text{ m/s} \end{gathered}

hence, the answer is 0.495 m/s

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