Question

1. A particle of mass m moves according to x = x0 + at2, y = bt3, and z = ct. Verify that

                                      dLdt=r×F=τ                      

Answer 1

Answer

Position vector can be written as

$r=x\hat {i}+y\hat {i}+z\hat {i}$

=$(x_0 + at^2) \hat {i} +bt^3 \hat {j}+ct \hat {k}$

Now acceleration can be written as

$a=\frac{d^2r}{dt^2}= (2a) \hat {i} +6bt\hat {j}$

Now cross product

$r×F=r×(ma)$

$=m(r×a)$

$=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}& \\ (x_0 + at^2) & bt^3 & ct \\ 2a & 6bt &0 \end{vmatrix}$

$=-6bct^2\hat{i}-2act \hat{j}+(6x_0bt+4abt^3) \hat{k}$

This means

$r×F ≠ 0$

In this specific case we can rewrite as

$dL/dt=r×F=τ$