Question

An object rotates about a fixed axis, and the angular position of a reference line on the object is given by θ = 0.220e3t, where θ is in radians and t is in seconds. Consider a point on the object that is 2.00 cm from the axis of rotation. At t = 0, what are the magnitudes of the point's (a) tangential component of acceleration and (b) radial component of acceleration?

Answer 1

Answer:
`a_(t)=3.96`

`a_(c)=0.8712`

Step-by-step explanation:

Given

`\theta =0.220e^(3t)`

r=2cm

Now angular velocity is given by
`\omega =\frac{\mathrm{d}\theta}{\mathrm{d}t}`

`\omega =0.66e^(3t)`

Now linear velocity(v) is given =
`\omega r`

`v=1.32e^(3t)`

Now tangential component of acceleration is given by

`a_(t)=\frac{\mathrm{d}\vec{v}}{\mathrm{d}t}=3.96e^(3t)`

at t=0

`a_(t)=3.96cm/s^2`

radial component of acceleration is given by

`a_(c)=\omega ^(2)r`

`a_(c)=0.4356e^(6t)* 2`

`a_(c)=0.8712e^(6t) cm/s^(2)`

at t=0

`a_c=0.8712 cm/s^(2)`