20.1 Angular mometum (check your understanding)  (Page 2/2)

From the alternatives given, it is evident that we need to find kinetic energy in terms of linear and angular momentum. Now, the kinetic energy of the particle is expressed as :

$$\begin{array}{l}K=\frac{1}{2}xm{v}^2\end{array}$$

The magnitude of linear momentum is given by :

$$\begin{array}{l}p=mv\end{array}$$

Squaring both sides, we have :

$$\begin{array}{l}\Rightarrow p^2=m^2{v}^2\end{array}$$

Rearranging,

$$\begin{array}{l}\Rightarrow m{v}^2=\frac{p^2}{m}\end{array}$$

Combining equations,

$$\begin{array}{l}K=\frac{p^2}{2m}\end{array}$$

Now, the magnitude of angular momentum is given by :

$$\begin{array}{l}\ell =mvr\end{array}$$

Squaring both sides, we have :

$$\begin{array}{l}\Rightarrow {\ell }^2=m^2{v}^2{r}^2\end{array}$$

$$\begin{array}{l}\Rightarrow m{v}^2=\frac{{\ell }^2}{m{r}^2}\end{array}$$

Combining equations,

$$\begin{array}{l}K=\frac{{\ell }^2}{2m{r}^2}\end{array}$$

Hence, options (a) and (c) are correct.

In order to find dimensional formula, we use the expression of angular momentum :

$$\begin{array}{l}\ell =mvr\sin \theta \end{array}$$

The dimension of angular momentum is :

$$\begin{array}{l}\left[K\right]=\left[m\right]\left[v\right]\left[r\right]\left[\sin \theta \right]=\left[M\right]\left[L{T}^{\mathrm{-}1}\right]\left[L\right]=\left[M{L}^2{T}^{\mathrm{-}1}\right]\end{array}$$

Hence, option (c) is correct.

This is a case of rotation. We can solve this problem using either the general expression or specific expression involving moment of inertia. When we use general expression, we should measure moment arm from the axis of rotation.

Since rod is rotating in anticlockwise direction, angular momentum of each particle is positive.

The angular momentum of first particle of mass “m” is :

$$\begin{array}{l}{\ell }_1=mvr=m\omega x\frac{2d}{3}x\frac{2d}{3}=\frac{4m\omega d^2}{9}\end{array}$$

The angular momentum of second particle of mass “2m” is :

$$\begin{array}{l}{\ell }_2=mvr=2m\omega x\frac{d}{3}x\frac{d}{3}=\frac{2m\omega d^2}{9}\end{array}$$

$$\begin{array}{l}\ell ={\ell }_1+{\ell }_2=\frac{4m\omega d^2}{9}+\frac{2m\omega d^2}{9}=\frac{2m\omega d^2}{3}\end{array}$$

Hence, option (b) is correct.

The angular momentum of a particle about a point is defined as :

$$\begin{array}{l}\mathbf{\ell }=m(\mathbf{r}\mathbf{x}\mathbf{v})\end{array}$$

The magnitude of angular momentum can be evaluated using any of the three methods discussed in the module on the subject. In this case, evaluation of angular momentum involving moment arm is most appropriate. Thus,

$$\begin{array}{l}\ell =m{r}_{\perp }v\end{array}$$

We, now, analyze the situation by drawing the figure. The path of the motion is shown. The moment arm i.e. the perpendicular distance of velocity vector from the origin is :

Angular momentum of a particle

$$\begin{array}{l}{r}_{\perp }=\mathrm{OB}=2\sin 45°=\surd 2\end{array}$$

Hence,

$$\begin{array}{l}\Rightarrow \ell =0.1x\surd 2x\surd 2=0.2\frac{\mathrm{kg}-m^2}{s}\end{array}$$

Hence, option (b) is correct.

We need to find a general equation for angular momentum for a given time “t”. We observe here that projectile motion is a two dimensional motion for which we can have information in component form. It is, therefore, required to obtain component information for position and the velocity.

Projectile motion

$$\begin{array}{l}x=u\cos \theta xt\\ y=u\sin \theta xt-\frac{1}{2}g{t}^2\\ v_x=u\cos \theta \\ v_y=u\sin \theta -gt\end{array}$$

Now, the angular momentum is :

ℓ = | i j k | | utcosθ utsinθ – 1/2g t*2 0 || u cosθ usinθ – gt 0 |

$$\begin{array}{l}\Rightarrow \ell =m\{u\cos \theta xt(u\sin \theta -gt)-u\cos \theta (u\sin \theta -\frac{1}{2}xg{t}^2\left)\right\}\mathbf{k}\end{array}$$

$$\begin{array}{l}\Rightarrow \ell =m(u\cos \theta xtxu\sin \theta -u\cos \theta xtxgt-u\cos \theta xu\sin \theta xt+\frac{1}{2}xg{t}^{2}xu\cos \theta )\mathbf{k}\end{array}$$

$$\begin{array}{l}\Rightarrow \ell =m(u^{2}t\cos \theta x\sin \theta -ug{t}^2\cos \theta -u^{2}t\cos \theta x\sin \theta +\frac{1}{2}xug{t}^2\cos \theta )\mathbf{k}\end{array}$$

$$\begin{array}{l}\Rightarrow \ell =-\frac{1}{2}xmug{t}^2\cos \theta \mathbf{k}\end{array}$$

Hence, option (d) is correct.