The position of a particle is given i(t) = A(coswt i + sin wt 1), where w is a constant. (a) Show that the particle moves in a circle of rad

Question

The position of a particle is given i(t) = A(coswt i + sin wt 1), where w is a constant. (a) Show that the particle moves in a circle of radius A. (b) Calculate and then show that the speed of the particle is a constanta (c) Determine and show that a is given by ae = rw?. (d) Calculate the centripetal force on the particle.

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Phúc Điền 3 months 2021-08-13T21:03:35+00:00 1 Answers 0 views 0

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    2021-08-13T21:04:46+00:00

    Answer:

    Explanation:

    (a) we can prove that the particle moves in a circle by taking the square of the norm of r(t)

    |r(t)|=\sqrt{A^{2}[cos^{2}(\omega t)+sin^{2}(\omega t)]}\\cos^{2}(\omega t)+sin^{2}(\omega t)=1\\|r(t)|=A\\r=A

    the norm of the position vector does not depend of time, so |r| is constant and is a radius of a circle.

    (b) the sped of the particle is the norm of the velocity v(t). Velocity is calculated by derivating r(t)

    v(t)=\frac{dr(t)}{dt}=A(-\omega sin(\omega t)\hat{i}+\omega cos(\omega t)\hat{j})\\|v(t)|=\sqrt{A^{2}\omega^{2}(sin^{2}(\omega t)+cos^{2}(\omega t))}\\v=A\omega

    A and w are constant, hence the speed of the particle is constant.

    (c) the acceleration is the derivative of the velocity

    a(t)=\frac{dv(t)}{dt}=A(-\omega ^{2}cos(\omega t)\hat{i}-\omega^{2}sin(\omega t)\hat{j})\\ a(t)=-\omega^{2}r(t)\\|a(t)|=a=\omega^{2}r

    (d)

    F_{c}=ma_{c}=m\frac{v^{2}}{r}=m\frac{A^{2}\omega ^{2}}{A}=mA\omega ^{2}

    I hope this is useful for you

    regards

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