Question #154669
  1. A particle of mass m moves according to x = x0 + at2, y = bt3, and z = ct. Verify that


                                      dLdt=r×F=τ                      


1
Expert's answer
2021-01-11T11:42:06-0500

Answer

Position vector can be written as

r=xi^+yi^+zi^r=x\hat {i}+y\hat {i}+z\hat {i}

=(x0+at2)i^+bt3j^+ctk^(x_0 + at^2) \hat {i} +bt^3 \hat {j}+ct \hat {k}

Now acceleration can be written as

a=d2rdt2=(2a)i^+6btj^a=\frac{d^2r}{dt^2}= (2a) \hat {i} +6bt\hat {j}

Now cross product

r×F=r×(ma)r×F=r×(ma)

=m(r×a)=m(r×a)

=i^j^k^(x0+at2)bt3ct2a6bt0=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}& \\ (x_0 + at^2) & bt^3 & ct \\ 2a & 6bt &0 \end{vmatrix}

=6bct2i^2actj^+(6x0bt+4abt3)k^=-6bct^2\hat{i}-2act \hat{j}+(6x_0bt+4abt^3) \hat{k}

This means

r×F0r×F ≠ 0

In this specific case we can rewrite as

dL/dt=r×F=τdL/dt=r×F=τ


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