Question

The path of motion of a 8-lb particle in the horizontal plane is described in terms of polar coordinates as r = (2 t + 1)ft and θ = (0.2 t2 − t) rad, where t is in seconds.

Answer 1

Missing question in the text:
"Determine the magnitude of the resultant force acting on the particle when t=2s."

For Newton's second law, the force is given by

`F=ma`
where m is the mass of the particle and a its acceleration.

The acceleration in polar coordinates is given by:

`a= \sqrt{a_r^2+a_(\theta)^2}`
where

`a_r = r''-r \theta'^2`

`a_(\theta) = r \theta'' +2r'\theta '`
are the radial and tangential accelerations.

Therefore, to find the acceleration starting from the polar coordinates, we need to find the derivatives of r and
`\theta`. Let's compute them:

`r=2t+1`

`r'=2`

`r''=0`
We need the values at t=2s, therefore we have

`r=5 ft=1.52 m`

`r'=2 ft=0.61 m`

`r''=0 ft = 0 m`

and

`\theta = 0.2 t^2-t`

`\theta' = 0.4t-1`

`\theta''=0.4`
And using t=2s,

`\theta = -1.2 rad`

`\theta' = -0.2 rad`

`\theta''=0.4 rad`

And now we can calculate the acceleration:

`a_r=r''-r\theta '^2 =-0.06`

`a_(\theta)=r\theta ''+2r' \theta ' = 0.36`
and

`a= \sqrt{a_r^2+a_(\theta)^2}= √((-0.06)^2+(0.36)^2)=0.35 m/s^2`
And using the mass:

`m=8 lb=3.63 kg`
we find the force:

`F=ma=3.63 kg\cdot 0.35 m/s^2=1.29 N`