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(D) The torsional rigidity of a rod is given by $C = \frac{\pi \eta r^4}{2L}$,where $\eta$ is the modulus of rigidity. Since the rods are joined in se

Vedclass · Accepted Answer

$8	heta /9$

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(D) The torsional rigidity of a rod is given by $C = \frac{\pi \eta r^4}{2L}$,where $\eta$ is the modulus of rigidity. Since the rods are joined in series,the torque $	au$ transmitted through both rods is the same. Let $	heta_0$ be the twist angle at the joint. For the small rod (length $l/2$,radius $r/2$): $	au = C_1(	heta_0 - 0) = \frac{\pi \eta (r/2)^4}{2(l/2)} 	heta_0 = \frac{\pi \eta r^4}{16(l/2)} 	heta_0 = \frac{\pi \eta r^4}{8l} 	heta_0$. For the large rod (length $l$,radius $r$): $	au = C_2(	heta - 	heta_0) = \frac{\pi \eta r^4}{2l} (	heta - 	heta_0)$. Equating the torques: $\frac{\pi \eta r^4}{8l} 	heta_0 = \frac{\pi \eta r^4}{2l} (	heta - 	heta_0)$. $\frac{	heta_0}{8} = \frac{	heta - 	heta_0}{2} \Rightarrow 	heta_0 = 4(	heta - 	heta_0) \Rightarrow 	heta_0 = 4	heta - 4	heta_0 \Rightarrow 5	heta_0 = 4	heta \Rightarrow 	heta_0 = 4	heta/5$. Wait,re-evaluating the expression: $\frac{	heta_0}{4} = 	heta - 	heta_0 \Rightarrow 	heta_0 = 4	heta - 4	heta_0 \Rightarrow 5	heta_0 = 4	heta$. Let's re-check the torsional constant: $C = \frac{\pi \eta r^4}{2L}$. $C_1 = \frac{\pi \eta (r/2)^4}{2(l/2)} = \frac{\pi \eta r^4 / 16}{l} = \frac{\pi \eta r^4}{16l}$. $C_2 = \frac{\pi \eta r^4}{2l}$. $C_1 	heta_0 = C_2(	heta - 	heta_0) \Rightarrow \frac{\pi \eta r^4}{16l} 	heta_0 = \frac{\pi \eta r^4}{2l} (	heta - 	heta_0) \Rightarrow \frac{	heta_0}{8} = 	heta - 	heta_0 \Rightarrow 	heta_0 = 8	heta - 8	heta_0 \Rightarrow 9	heta_0 = 8	heta \Rightarrow 	heta_0 = 8	heta/9$.

$A$ rod of length $l$ and radius $r$ is joined to a rod of length $l/2$ and radius $r/2$ of the same material. The free end of the small rod is fixed to a rigid base,and the free end of the larger rod is given a twist of $\theta^\circ$. The twist angle at the joint will be:

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