1. For the spiral r=e^Θ and Θ=t:

(a) compute the tangential & normal components of the velocity when t=0.

(b) compute the radial & angular components of the acceleration when t=0.

2. A particle moves along the ellipse r=30/(4+cosΘ) in an inverse square gravity field centered at the origin, and it is known that dΘ/dt = 1 when Θ=0.

(a) compute the particle's fastest & slowest speeds in its orbit.

(b) what is the period of the orbit?

Expert Answers

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The figure with the spiral is below.

`r = r^Theta` with `Theta=t`

The parametric equations of the spiral components `x, y` are

`x(t) =r*cos(alpha) =e^t*cos(omega*t)`

`y(t) =r*sin(alpha) =e^t*sin(omega*t)`

with the initial condition: `t=0 rArr Theta=0 => omega*t =0`

and `omega=constant`

In rectangular coordinates written as a vector we have

`r =x*hati +y*hatj =[e^t*cos(omega*t)]*hati +[e^t*sin(omega*t)]*hatj `

a)

The speed of the particle is by definition

`v =(dr)/(dt)`

which gives by components

`(dx)/dt =e^t*cos(omega*t) -omega*e^t*sin(omega*t) =x-omega*y`

`(dy)/dt =e^t*sin(omega*t) +omega*e^t*cos(omega*t) =y +omega*x`

Therefore the speed written as a vector is

`v = (x-omega*y)*hati +(y+omega*x)*hatj`

Since at `t=0` the radius `r` is horizontal, the normal speed is directed in the `hati` (x axis) vector direction, and the tangential speed is directed in the `hatj` (y axis) vector direction

`v_t(t=0) =y(0)+omega*x(0) =omega*e^0 =omega`

`v_n(t=0) =x(0)-omega*y(0) =e^0 =1`

b)

The acceleration is

`a (=(d^2r)/dt^2)=(dv)/dt`

On its components:

`(dv_x)/dt =d/dt(x-omega*y) =(x-omega*y)-omega(y+omega*x) =x-2*omega*y -omega^2*x`

`(dv_y)/dt^2 =d/dt(y+omega*x) =(y+omega*x) +omega(x-omega*y) =y+2*omega*x-omega^2*y`

Therefore the acceleration written as a vector is

`a =(x-2omega*y-omega^2*x)*hati +(y+2omega*x-omega^2*y)*hatj`

At `t=0` the normal component of `a` is in the `hati` (x axis) direction and the tangential component of `a` is in the `hatj` (y axis) direction.

`a_t(0) =y(0)+2*omega*x(0) -omega^2*y(0) =2*omega*e^0 =2*omega`

`a_n(0) =x(0)-2*omega*y(0)-omega^2*x(0) =1-omega^2`

Answer: the tangential and normal components of the speed, respectively acceleration at `t=0` are `omega` and 1, respectively `2*omega` and `1-omega^2`

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1. a) The tangential component of velocity is given by

`v_t = r* (d(theta))/(dt)`

Since `theta = t` , `(d(theta))/dt = 1` and when t = 0 `theta = 0` and r = 1. Thus, when

t = 0, `v_t = 1`  .

The radial component is given by

`v_r = (dr)/dt` . It can be found using chain rule:

`(dr)/(dt) = e^(theta) * (d(theta))/(dt) =1*1 = 1` at t = 0.

b) The radial component of acceleration is given by

`a_r = (d^2r)/(dt^2) - r((d(theta))/(dt))^2`

`(d^2r)/(dt^2) = d/dt (e^theta * (d(theta))/(dt)) =`

`=e^theta * ((d(theta))/(dt))^2 + e^theta * (d^2(theta))/(dt)^2`

`` Pluggin this expression in the one for `a_r` and considering that the second derivative of `theta` with respect to time is 0, and that `r = e^theta` ,

we obtain that `a_r = 0` .

The tangential acceleration is given by

`a_t = r*(d^2(theta))/(dt)^2 + 2 (dr)/(dt) * (d(theta))/(dt)` .

Again, considering that `(d^2(theta))/(dt)^2 = 0` , `(dr)/(dt) = 1` when t = 0 (see the result above) and `(d(theta))/(dt) = 1` , we get

`a_t = 2` .

Therefore, at t = 0:

`v_r = 1` , `v_t = 1` , `a_r = 0` and `a_t = 2` .

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