HSC Science (Computer Science) 12th Board ExamMaharashtra State Board
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In Circular Motion, Assuming V = W * R , Obtain an Expression for the Resultant Acceleration of a Particle in Terms of Tangential and Radial Component. - HSC Science (Computer Science) 12th Board Exam - Physics

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Question

In circular motion, assuming `bar v = bar w xx bar r` , obtain an expression for the resultant acceleration of a particle in terms of tangential and radial component.

Solution

Acceleration of a particle,

`a=lim_(deltat->0) ((delta v)/(delta t)) ... delta t ->0; delta t ne 0`

`therefore a=(dv)/(dt)`

`But, v = r omega`

`therefore a=d/dt (r omega)`

`=r (d omega)/dt + omega (dr)/(dt)`

`because "r is constant"`

`(dr)/(dt) = 0`

`therefore a=r (d omega)/(dt)`

`because (d omega)/(dt) = alpha`

`alpha = r alpha`

Given that:

`bar "v"= bar omega xx bar "r"`

Differentiating w.r.t. time

`(d bar v)/(dt)=d/dt (bar omega xx bar "r")`

`(d barv)/(dt) = (d baromega)/(dt) xx barr + baromega xx (d barr)/(dt)`

`(d barv)/(dt) = baralpha xx barr + baromega xx barv`

`therefore bara = bara_T + bara_r`

Where,

a = Linear acceleration

aT = Tangential component of linear acceleration

ar = Radial component of linear acceleration

  Is there an error in this question or solution?

APPEARS IN

 2014-2015 (March) (with solutions)
Question 2.1 | 2.00 marks

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Solution In Circular Motion, Assuming V = W * R , Obtain an Expression for the Resultant Acceleration of a Particle in Terms of Tangential and Radial Component. Concept: Radial Acceleration.
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