Question

A particle moves along a circular path of radius 10 ft. If its position is 8 = 0.74 - 1 rad, where t, is in seconds, determine the magnitude of the particle's acceleration when 8 = 0.53 radians. Note that the magnitude is a scalar, if you were asked for acceleration you would need to give a vector Give answer to at least 3 decimal places jäl =

Answer 1

Particle position is given by rad which is along a circle of radius 10 ft.

0 = 0.7 - 1

Acceleration which is here angular  acceleration which is given by :
A= Ré
and centripetal acceleration(acceleration responsible for circular motion) which is given by
A = R02

Please note this centripetal acceleration is directed radially inwards and angular acceleration is directed along the tangent to the circle.   

Now evaluating  

       

8 = 0.7 0.7

t ....?

Now determining time for

E9'0 = 0

  

0.53 = 0.7 -1

  

log(0.53 +1) = 0.77

  

t = log(0.53+1) 0.7

= 0.6075 sec

Now

8 = 0.74 0.740.0075

= 0.7497 rad / s^2

So

A| = |RÖ = 10 * 0.7497

= 7.497 ft /s ^2

Centripetal acceleration :

0 = 0.700.77

= 1.071 rad /sec

and

A2 = R02

= 11.47 ft/s^2

since acceleration are orthogonal

the net acceleration is given by

Anet) = V A2 + A

= 13.703 ft/s^2

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