Question

The three-dimensional motion of a particle on the surface of a right circular cylinder is described by the relations r = 2 (m) θ = πt (rad) z = sin24θ (m) Compute the velocity and acceleration of the particle at t=5 s.

Answer 1

`V_(rex)=75.65m/s` and
`a_(res)=0` at t=5 secs

Step-by-step explanation:

We have r =2m

`\therefore (dr)/(dt)=0\\\\=>V_(r)=0`

Similarly

`=>V_(\theta )=\omega r\\\\\omega =(d\theta )/(dt)=(d(\pi t))/(dt)=\pi \\\\\therefore V_(\theta )=\pi r=2\pi`

Similarly

`=>V_(z )=(dz)/(dt)\\\\V_(z)=(dsin(24\pi t))/(dt)\\\\V_(z)=24\pi cos(24\pi t)`

Hence

at t =5s
`V_(\theta)=2\pi m/s`

`V_(z)=24\pi cos(120\pi)`

`V_(z)=24\pi m/s`

`V_(res)=\sqrt{V_(\theta )^(2)+V_(z)^(2)}`

Applying values we get

`V_(res)=75.65m/s`

Similarly

`a_(\theta )=(dV_(\theta ))/(dt)=(d(2\pi) )/(dt)=0\\\\a_(z)=(d^(2)(sin(24\pi t)))/(dt^(2))\\\\a_(z)=-24^(2)\pi^(2)sin(24\pi t)\\\\\therefore t=5\\a_(z)=0`