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A particle is moving in a circular orbit...

A particle is moving in a circular orbit when a constant tangential acceleration. After `2s` from the beginning of motion, angle between the total acceleration vector and the radius E becomes `45^(@)`. What is the angular acceleration of the particle?

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In the adjoining figure are shown the total acceleration vector `vec(a)` and its components the tangential accelerations `vec(a)_(tau)` and normal acceleration `vec(a)_(n)` are shown. These two components are always mutually perpendicular to each other and act along the tangent to the circle and radius respectively. Therefore, if the total acceleration vector makes an angle of `45^(@)` with the radius, both the tangential and the normal components must be equal in magnitudr.
Now from equation and, we have
`a_(tau)=a_(n) rarr alpha R=omega^(2)R rArr alpha=omega^(2)` ...(i)
Since angular acceleration is uniform, form equation, we have `omega=omega_(o)+alphat`
Substituting `omega_(0)=0` and `t=2 s`, we have `omega=2alpha` ...(ii)
From equation (i) and (ii), we have `alpha=0.25 rad//s^(2)`
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Knowledge Check

  • A particle of mass 10 g moves along a circle of radius 6.4 cm with constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal 8xx10^(-4)J by the end of the second revolution after the beginning of the motion?

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    B
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    C
    `0.2m//s^(2)`
    D
    `0.1m//s^(2)`

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