Question

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

Answer 1

Restoring torque=c theta=πSr4( theta-thetao)/2l=constant

Answer 2

Given that,

Radius = r

Length = l

Radius of another rod $r'=\dfrac{r}{2}$

Length of another length $l'=\dfrac{l}{2}$

We need to calculate the twist angle at the joint

Using formula of torque

$\tau=c\theta$

Put the value into the formula

$\tau=\dfrac{\pi nr^4\theta}{2l}=constant$

For both rod,

$\dfrac{\pi nr^4(\theta-\theta_{0})}{2l}=\dfrac{\pi n(\dfrac{r}{2})^4(\theta_{0}-0)}{2(\dfrac{l}{2})}$

$\dfrac{\theta-\theta_{0}}{2l}=\dfrac{\theta_{0}}{16l}$

$\dfrac{\theta-\theta_{0}}{2}=\dfrac{\theta_{0}}{16}$

$\theta=\dfrac{9\theta_{0}}{8}$

Hence, The twist angle at the joint is $\dfrac{9\theta_{0}}{8}$