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Laws Of Motion

Question
CBSEENPH11016974

Derive the relation between linear acceleration and angular acceleration.

Solution

Let a particle be revolving in a circle of radius r.
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#6 {main}</pre> be its position vector and <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre> be its angular velocity at any instant.
The linear velocity of particle at that instant is, 
                   straight v with rightwards harpoon with barb upwards on top space equals space omega with rightwards harpoon with barb upwards on top space x space r with rightwards harpoon with barb upwards on top

The linear acceleration of object is,
                             
straight alpha with rightwards arrow on top space equals space fraction numerator straight d straight v with rightwards arrow on top over denominator dt end fraction equals fraction numerator straight d left parenthesis straight omega with rightwards arrow on top cross times straight r with rightwards arrow on top right parenthesis over denominator dt end fraction 
   equals straight omega with rightwards arrow on top cross times fraction numerator straight d straight r with rightwards arrow on top over denominator dt end fraction plus fraction numerator straight d straight omega with rightwards arrow on top over denominator dt end fraction cross times straight r with rightwards arrow on top 
   equals space omega with rightwards arrow on top cross times v with rightwards arrow on top plus alpha with rightwards arrow on top cross times r with rightwards arrow on top 
   equals space stack alpha subscript r with rightwards arrow on top plus stack alpha subscript 1 with rightwards arrow on top 
The magnitude of net acceleration of particle is,
straight a space equals space square root of straight a subscript straight t squared plus space straight a subscript straight r squared end root space

Here comma space we space have

straight omega with rightwards harpoon with barb upwards on top space perpendicular space straight v with rightwards harpoon with barb upwards on top space and space straight alpha with rightwards harpoon with barb upwards on top space perpendicular space straight r with rightwards harpoon with barb upwards on top space

therefore space open vertical bar straight omega with rightwards harpoon with barb upwards on top space perpendicular space straight v with rightwards harpoon with barb upwards on top close vertical bar space equals space straight omega space straight v space comma space and space

open vertical bar straight alpha with rightwards harpoon with barb upwards on top space straight x stack space straight r with rightwards harpoon with barb upwards on top close vertical bar space equals space straight alpha space straight r space

Now comma space

straight alpha space equals space square root of left parenthesis ωv right parenthesis squared space plus space left parenthesis αr right parenthesis squared end root space

space space space equals space square root of open parentheses straight omega squared straight r close parentheses squared space plus space left parenthesis αr right parenthesis squared space end root space
               
Total acceleration makes an angle θ with radial acceleration and is given by, 
tan space straight theta space equals space straight alpha subscript straight t over straight alpha subscript straight r space equals space fraction numerator αr over denominator straight omega squared straight r end fraction space equals space straight alpha over straight omega squared space