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Derve the relation between tangential acceleration and angular acceleration. |
Answer» Solution :Consider an object moving along a circle of radius r. In a time `Deltat`, the object travels in an are distance As as shown in figure. The corresponding angle subtended is `Deltatheta`. The `Deltas` can be written in terms of `Deltatheta` as,  `Deltas=rDeltatheta""......(1)` In a time `Deltat`, we have `(Deltas)/(Deltat)=t""(Deltatheta)/(Deltat)""......(2)` In the LIMIT `Deltat"to0`, the above equation becomes `(ds)/(dt)=romega"".......(3)` Here `(ds)/(dt)` is linear speed (5) which is tangential to the circle and ois angular speed. So equation (3) becomes `v_(r)=romega""......(4)` which gives the relation between linear speed and angular speed. Eq. (4) is true only for circular motion. In general the relation between linear and angular velocity is GIVEN by `vecv=vecomegaxxvecr"".......(5)` For circular motion eq. (5) reduces to eq. (4) since `vecomegaandvecr` are perpendicular to each other. Differentiating the eq. (4) with respect to time, we get (since r is constant) `(DV)/(dt)=(rdomega)/(dt)=ralpha` Here `(dv)/(dt)` is the tangential acceleration and is denoted as `a_(t)=(domega)/(dt)` is the angular acceleration `alpha`. Then eq. (5) becomes `a_(t)=ralpha""......(6)`. |
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