A rigid body rotates about an axis through the origin with angular velocity $10\sqrt{3} $ rad/sec. If $\overrightarrow{\omega}$ points in the direction of $\ \hat{i} + \hat j + \hat k$, then what is the equation to the locus of the points having tangential speed $20\ $ m/sec?
I have no idea about this problem. Please help.
If a rigid body rotates around an axis with angular velocity $\overrightarrow \omega$ then : $$\overrightarrow \omega = |\omega|\space \widehat \omega\space \text{rad/sec}$$ where $|\omega|$ is the magnitude and $\widehat \omega$ is the direction of rotation or the axis of rotation. According to the problem, $$\overrightarrow \omega = 10 \sqrt{3} \space \frac{(\widehat i + \widehat j + \widehat k)}{(\sqrt{3})}\space \text{rad/sec} = 10 (\widehat i + \widehat j + \widehat k)\space \text{rad/sec} $$ Now the particle rotates around this axis with a tangential speed of $v_T=20\space m/sec$, so the perpendicular distance between the axis of rotation and the axis is : $$v_T=|\omega|r$$ giving $r=\frac{2}{\sqrt{3}}m$. The locus of a point having a tangential speed of $v_T$ and rotating around the axis is definitely a circle in 2 D and a cylinder in 3 D.
To find the equation of cylinder, the axis given is the line passing through the origin(mentioned in the question), so find the equation of a line which passes through $(0,0,0)$ and is in the direction of $\widehat i + \widehat j + \widehat k$. Find the direction cosines and apply the equation of line. Such a line is : $$\frac{x-0}{\frac{1}{\sqrt{3}}}=\frac{y-0}{\frac{1}{\sqrt{3}}}=\frac{z-0}{\frac{1}{\sqrt{3}}}=t$$ Now consider a point $P(h,k,l)$ which is the rigid body location, this point is always transalting a path in which the point is always maintaing a tangential speed of $v_T=20 \space m/sec$ The trajectory of all such points is the locus. If we visualize it, we see the set of all points that are at a particular distance from the rotation axis line.
To find the equation of the cylinder just apply the fact that the line joining the axis and the point $P$ are perpendicular to each other and at a distance of $\frac{2}{\sqrt3} m$.
Direction cosines of the lines OX and HX will follow the relation:
$$t(h-t)+t(k-t)+t(l-t)=0$$
so either $t=0$ or $$h+k+l=3t$$
Now the distance HX = $\frac{2}{\sqrt 3}$. So:
$$HX^2=(h-t)^2+(k-t)^2+(l-t)^2=\frac{4}{3}$$
From Pythagoras theorem, $OX^2+HX^2=OH^2$
Solving the last two equations you can get:
$$h^2+k^2+l^2=\frac{4}{3}+3t^2$$
Now put back the value of $t$ as $\frac{h+k+l}{3}$ to get:
$$h^2+k^2+l^2-hk-kl-lh=2$$ Hope this helps...