Figuring out velocity,acceleration, work of a particle given that we know its position vector.

Recently this question came up in a problem class of mine.

A particle moves in such a way that its position vector at any time $t$ is $\vec{r}(t)=\pmatrix{A\sin{\omega t}\\A\cos{\omega t}\\Bt^2}$, where $A$, $B$ and $\omega$ are known positive constants.

(i) Determine the velocity $\vec{v}(t)$.

(ii) Determine the speed $v(t)$.

(iii) Calculate the acceleration $\vec{a}(t)$ and give its components that are tangential and normal to the trajectory of the particle.

(iv) If $m$ is the mass of the particle what force $\vec{F}(t)$ acts on the particle and what work is performed by this force during the time interval from $t_0 = 0$ to $t_1 = \pi/\omega$?

Remarks:For i)The velocity is just the derivative of the position vector,right?

ii) The speed is the magnitude of the velocity vector, but how would I right it out as I don't have it for a particular moment in time.

iii)The acceleration is the 2nd derivative of the position vector but how do we find the the tangental and normal components given that we know the tangent unit vector to the trajectory of the particle is $(1/v(t))\cdot\vec{v}(t)$.

iv)To be honest really not sure how to work this one out.

Any help is appreciated, thanks

• You have r and $\theta$ as function of time. Use the derivatives to get velocity and acceleration. Use dot multiplications to determine perpendicularity or parallelizem. – Moti May 7 '16 at 16:55
• Once you have the acceleration, force is trivial. Work is given by the integral $\int_\Gamma F\cdot ds$, so you’ll be using the tangential component of the acceleration that you computed in part (iii). – amd May 7 '16 at 19:37

$$\vec{r}(t)=\pmatrix{A\sin{\omega t}\\A\cos{\omega t}\\Bt^2}\Rightarrow \vec{v}(t)=\vec{r}\,'(t)=\pmatrix{A\omega\cos{\omega t}\\-A\omega\sin{\omega t}\\2Bt}\Rightarrow \vec{a}(t)=\vec{v}\,'(t)=\pmatrix{-A\omega^2\sin{\omega t}\\-A\omega^2\cos{\omega t}\\2B}$$ So we have $$v(t)=| \vec{v}(t)|=\sqrt{(A\omega\cos{\omega t})^2+(-A\omega\sin{\omega t})^2+(2Bt)^2}=\sqrt{A^2\omega^2+4B^2t^2}$$ and $$a(t)=| \vec{a}(t)|=\sqrt{(-A\omega^2\sin{\omega t})^2+(-A\omega^2\cos{\omega t})^2+(2B)^2}=\sqrt{A^2\omega^4+4B^2}$$ We know that $\vec{v}(t)=v(t)\frac{\vec{v}(t)}{{v}(t)}=v(t)\hat{T}$ and $\vec{a}(t)=a_N(t)\hat{N}+a_T(t)\hat{T}$ where $$a_T(t)=v'(t)=\frac{4B^2t}{\sqrt{A^2\omega^2+4B^2t^2}}$$ and $$a_N(t)=\sqrt{a^2(t)-a_T^2(t)}=\sqrt{{A^2\omega^4+4B^2-\frac{16B^4t^2}{A^2\omega^2+4B^2t^2}}}$$ Thus the force is $$\vec{F}(t)=m\vec{a}(t)=\pmatrix{-mA\omega^2\sin{\omega t}\\-mA\omega^2\cos{\omega t}\\m2B}=ma_N(t)\hat{N}+ma_T(t)\hat{T}=F_N(t)\hat{N}+F_T(t)\hat{T}$$ The infinitesimal work is $\mathrm d W=\mathrm d\vec{F}(t)\cdot \mathrm d\vec{r}(t)=\mathrm d\vec{F}(t)\cdot \vec{v}(t)\mathrm dt= F_T(t)v(t)\mathrm dt$ and then $$W=\int_{t_0}^{t_1}\mathrm d W=\int_{t_0}^{t_1}F_T(t)v(t)\mathrm dt=\int_{0}^{\pi/\omega}m4B^2t\,\mathrm dt=\frac{m2B^2\pi^2}{\omega^2}$$