Chapter 11: Q8Q (page 320)

Figure 11-27 shows an overheadview of a rectangular slab that can
spin like a merry-go-round about its center at O. Also shown are seven
paths along which wads of bubble gum can be thrown (all with the
same speed and mass) to stick onto the stationary slab. (a) Rank the paths according to the angular speed that the slab (and gum) will have after the gum sticks, greatest first. (b) For which paths will the angular momentum of the slab(and gum) about Obe negative from the view of Fig. 11-27?

Short Answer

Expert verified

(a) Rank of the paths according to the angular speed that the slab will have after the gum sticks is . $r_{4} > r_{6} > r_{7} > r_{1} > r_{2} = r_{3} = r_{5} = 0$

(b) Path have the negative angular momentum of the slab about the center $O .$

Step by step solution

Step 1: Given

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

Determining the concept

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:

$\begin{matrix} \vec{L} = m v r \sin θ \\ = m (\vec{r} \times \vec{v}) \end{matrix}$

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

(a) Ranking the paths according to the angular speed that the slab (and gum) will have after the gum sticks

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,

$\vec{L} = m v r \sin θ$

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

The angle between $\vec{v}$ and $\vec{r}$ 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 $\vec{v}$ and $\vec{r}$ 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 $r_{4} > r_{6} > r_{7} > r_{1} > r_{2} = r_{3} = r_{5} = 0$ .

(b) Determining for which paths will the angular momentum of the slab (and gum) about centre 0 be negative

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 $r_{4} > r_{6} > r_{7} > r_{1} > r_{2} = r_{3} = r_{5} = 0$ .