1. The physical quantity angular momentum L is defined to be L = r × p, where r is the position vector and p is the momentum vector. Momentum is the product of mass and velocity. (... 1. The physical quantity angular momentum L is defined to be L = r × p, where r is the position vector and p is the momentum vector. Momentum is the product of mass and velocity. (a) Determine the dimension of angular momentum. (b) If the position vector of a 2.0 kg particle is r = (3t² i − 4t j + 2t² k) m, where t is in seconds, (i) find the velocity vector of the particle in terms of time, (ii) determine the particle’s position and momentum vectors at t = 0.50 s, and (iii) find its angular momentum vector at t = 0.50 s. (20 marks) Read more

Steps to Solve

1. Determine the dimension of angular momentum

Angular momentum ( L ) is defined as ( L = r \times p ), where ( r ) is the position vector and ( p ) is the momentum vector.

Momentum ( p ) is defined as the product of mass ( m ) and velocity ( v ): $$ p = mv $$

The dimension of mass ( m ) is ([M]), and the dimension of velocity ( v ) is ([\frac{L}{T}]).

Thus, the dimension of momentum ( p ) is: $$ [p] = [M] \cdot \left[\frac{L}{T}\right] = [M L T^{-1}] $$

Now, the dimension of angular momentum ( L ): Since ( r ) has dimensions ([L]), we have: $$ [L] = [L] \times [p] = [L] \times [M L T^{-1}] = [M L^2 T^{-1}] $$

2. Find the velocity vector of the particle

The position vector is given as: $$ \mathbf{r}(t) = (3t^2 , \hat{i} - 4t , \hat{j} + 2t^2 , \hat{k}) , m $$

To find the velocity vector ( \mathbf{v}(t) ), take the derivative of the position vector with respect to time ( t ): $$ \mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = \left( \frac{d}{dt}(3t^2) , \hat{i} + \frac{d}{dt}(-4t) , \hat{j} + \frac{d}{dt}(2t^2) , \hat{k} \right) $$

Calculating the derivatives: $$ \mathbf{v}(t) = (6t , \hat{i} - 4 , \hat{j} + 4t , \hat{k}) , m/s $$

3. Determine the particle’s position and momentum vectors at ( t = 0.50 , s )

First, substitute ( t = 0.50 , s ) into the position vector: $$ \mathbf{r}(0.50) = (3(0.50)^2 , \hat{i} - 4(0.50) , \hat{j} + 2(0.50)^2 , \hat{k}) $$

Perform the calculations:

• ( 3(0.50)^2 = 0.75 )
• ( -4(0.50) = -2.0 )
• ( 2(0.50)^2 = 0.5 )

Thus, $$ \mathbf{r}(0.50) = (0.75 , \hat{i} - 2.0 , \hat{j} + 0.5 , \hat{k}) , m $$

Next, find the velocity at ( t = 0.50 , s ): $$ \mathbf{v}(0.50) = (6(0.50) , \hat{i} - 4 , \hat{j} + 4(0.50) , \hat{k}) $$

Calculating the components:

• ( 6(0.50) = 3 )
• The remaining components are: ( -4 ), and ( 4(0.50) = 2 )

So, $$ \mathbf{v}(0.50) = (3 , \hat{i} - 4 , \hat{j} + 2 , \hat{k}) , m/s $$

Now, using ( m = 2.0 , kg ) to compute the momentum: $$ \mathbf{p} = m \mathbf{v}(0.50) = (2.0)(3 , \hat{i} - 4 , \hat{j} + 2 , \hat{k}) = (6 , \hat{i} - 8 , \hat{j} + 4 , \hat{k}) , kg \cdot m/s $$

4. Find angular momentum vector at ( t = 0.50 , s )

The angular momentum vector is given by: $$ \mathbf{L}(t) = \mathbf{r}(t) \times \mathbf{p} $$

Calculating the cross product using the position and momentum vectors calculated previously: $$ \mathbf{L}(0.5) = (0.75 , \hat{i} - 2.0 , \hat{j} + 0.5 , \hat{k}) \times (6 , \hat{i} - 8 , \hat{j} + 4 , \hat{k}) $$

Using the determinant method for the cross product: $$ \mathbf{L}(0.5) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0.75 & -2.0 & 0.5 \ 6 & -8 & 4 \end{vmatrix} $$

Calculating this determinant gives: $$ \mathbf{L}(0.5) = ((-2.0)(4) - (0.5)(-8)) \hat{i} - ((0.75)(4) - (0.5)(6)) \hat{j} + ((0.75)(-8) - (-2.0)(6)) \hat{k} $$

Which simplifies to: $$ \mathbf{L}(0.5) = (-8 + 4) \hat{i} - (3 - 3) \hat{j} + (-6 + 12) \hat{k} $$ $$ \mathbf{L}(0.5) = (-4 , \hat{i} + 0 , \hat{j} + 6 , \hat{k}) , kg \cdot m^2/s $$