91Ó°ÊÓ

The rotating analogue of linear momentum is angular momentum (also known as moment of momentum or rotational momentum). Because it is a conserved quantity—the total angular momentum of a closed system remains constant—it is a significant quantity in physics.

The product of the moment of inertia and angular velocity is angular momentum, which is an important feature of a rotating object.

The wheel's linear and angular velocity are depicted in the diagram below.

Here,$m$is mass,$v$is velocity, and$r$is the position vector.

Since the clay has the linear momentum before it touches the wheel and rotates with it, the linear momentum of the clay from the figure is given as,

$L_{clay}=mvR\cos 45°$

Here,$R$is the radius of the wheel. Negative sign indicates the direction.

Substitute $0.5kg$for $m,0.9m$and $5m/s$for $v$.

$L_{clay}=-\left(0.5kg\right)\left(5m/s\right)\left(0.9m\right)\cos 45°$

$=-1.5909kg·m^2/s$

The angular momentum of the wheel is given as,

$L_{wheel}=I\mathrm{Ó\neg }$

Here, $I$is moment of inertia and $\mathrm{Ó\neg }$is the angular velocity.

Moment of $I$inertia is expresses as,

$I=M{R}^2$

Here, $M$is the mass of the wheel and $R$is the radius.

Substitute $M{R}^2$for $I$in equation $L_{wheel}=-I\mathrm{Ó\neg }$.

$L_{wheel}=M{R}^2\mathrm{Ó\neg }$

Substitute $4.8kg$for localid="1668593889199" $M,0.9m$for $R$, and $0.33rad/s$for $\mathrm{Ó\neg }$.

$L_{wheel}=\left(4.8kg\right){\left(0.9m\right)}^2\left(0.33rad/s\right)$

$=1.2830kg·m^2/s$

Before the impact of the aforementioned estimated values, the initial angular momentum of the entire system is given as,

$L_i=L_{clay}+L_{wheel}$
$=1.5909kg·m^2/s+\left(1.2830kg·m^2/s\right)$

$=-0.323kg·m^2/s$

Therefore, the angular momentum of the whole system is $-0.323kg·m^2/s$and the direction is out of the page.

(b) The angular momentum of the clay just after the impact is given as,

$L_{clay}=m{r}^2\mathrm{Ó\neg }$

The final angular momentum of the whole system is given as,

$L_{system}=L_{clay}+L_{wheel}$

Substitute role="math" localid="1668594915315" $m{R}^2\mathrm{Ó\neg }$for $L_{wheel}$and $m{R}^2\mathrm{Ó\neg }$for $L_{clay}.$

$L_f=m{R}^2\mathrm{Ó\neg }+M{R}^2\mathrm{Ó\neg }$

$=\left(m+M\right)R^2\mathrm{Ó\neg }$

Apply the law of conservation of angular momentum by which the initial angular momentum before the impact and the final angular momentum after the impact is same.

$L_f=L_i$

$\left(m+M\right)R^2\mathrm{Ó\neg }=0.323kg·m^2/s$

Therefore, the final angular momentum of the system just after the impact is$-0.323kg·m^2/s$ and the direction is out of the page.

(c) Apply the law of conservation of energy to find out the angular velocity of the wheel after impact

$L_f=L_i$

$\left(m+M\right)R^2\mathrm{Ó\neg }=L_i$

Re-arrange the above equation to solve for $\mathrm{Ó\neg }$.

$\mathrm{Ó\neg }=\frac{L_i}{\left(m+M\right)R^2}$

Substitute $4.8kg$for $M$$0.9m$for $R$, and $-0.323kg·m^2/s$for $L_i$

$\mathrm{Ó\neg }=\frac{-0.323kg·m^2/s}{\left(0.5kg+4.8kg\right){\left(0.9m\right)}^2}$

$=-0.75rad/s$

Therefore, the angular velocity of the wheel after impact is$-0.75rad/s$.

(d) The linear momentum of combined system decreases because the axle exerts an upwards force.

Hence, the correct option is (1).