The man stands on a platform which is free to rotate. He has a rotating bicycle wheel in his hands. Consider platform + man + wheel as the system. The angular momentum of this system in a particular direction must be conserved if there is no net external torque in that direction. When the wheel is vertical initially, the direction of angular momentum is along the axis of the wheel. There is no component of angular momentum along vertical axis (the axis passing through the man). When the man changes the axis of the wheel making it vertical, the direction of angular momentum of the wheel becomes along the axis of the wheel. But, there can be no component of the angular momentum in that direction (no external torque in that direction). Hence, to cancel the vertical component of the angular momentum, man + platform start rotating.
A man starts walking from the point P (-3, 4), touches the x-axis at R, and then turns to reach at the point Q (0, 2). The man is walking at a constant speed. If the man reaches the point Q in the minimum time, then $50 [(PR)^2 + (RQ)^2 ]$ is equal to _ _ _ _ . Solution For time to be minimum at constant speed, the directions must be symmetric. In other words, the angles made by PR and RQ with the vertical must be the same just like in the law of reflection in optics. $tan \theta = \frac {MP}{MR} = \frac {NQ}{NR} $ $\Rightarrow \frac {3-r}{4} = \frac {r}{2}$ $\Rightarrow r=1 $ So, $R \equiv ( - 1,0)$ Now, $50(PR^2+RQ^2)=50[(4+16)+(1+4)]=1250$