91Ó°ÊÓ

Seven paths are shown along which wads of bubble gum can be thrown.

Using the equation of angular momentum, rank the paths according to the angular speed that the slab will have after the gum sticks, (greater first).

The formula is as follows:

$\begin{array}{c}\stackrel{→}{L}=mvr\sin \mathrm{θ}\\ =m(\stackrel{→}{r}×\stackrel{→}{v})\end{array}$

Where L is angular momentum, m is mass, r is a radius, I is a moment of inertia andvis velocity.

The gum-slab system spins about its center point O. The angular momentum of the gum-slab system about the point will remain constant ifthere isno external torque acting on it. There is no external torque acting on it, hence the initial angular momentum and the final angular momentum of the gum-slab remain constant.

To find initial angular momentum,

$\stackrel{→}{L}=mvr\sin \mathrm{θ}$

Here,$m\&v$remains constant, but$r$and$\mathrm{θ}$changes in all paths.

The angle between$\stackrel{→}{v}$and$\stackrel{→}{r}$is$0^0$for the paths$2,3\&5.$Hence, the magnitude of initial angular momentum of the bubble gum due to the path$2,3\&5$is zero.

$L_2=L_3=L_5=0$

The angle between$\stackrel{→}{v}$and$\stackrel{→}{r}$is${90}^0$for the paths.$1,4,6\&7$Hence, the magnitude of angularmomentum is determined by the magnitude of the position vector.

From figure,

$r_4>r_6>r_7>r_1$

Therefore, the rank of the initial angular momentum of the bubble gum is,

$L_4>L_6>L_7>L_1>L_2=L_3=L_5=0$

And the rank of the final angular momentum of the bubble gum is,

$L_4>L_6>L_7>L_1>L_2=L_3=L_5=0$

Hence, therank of the paths according to the angular speed that the slab will have after the gum sticks is$r_4>r_6>r_7>r_1>r_2=r_3=r_5=0$.

Path$1,4\&7$have the negative angular momentum of the slab about the centre$O.$

Hence, using the equation of angular momentum, the paths can be ranked according to the angular speed that the slab will have after the gum sticks, great first and the rank is $r_4>r_6>r_7>r_1>r_2=r_3=r_5=0$.