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A wheel rotates around a stationary axis so that the rotation angle `theta` varies with time as `theta=at^(2)` where `a=0.2rad//s^(2)`. Find the magnitude of net acceleration of the point A at the rim at the moment `t=2.5s` if the linear velocity of the point A at this moment is `v=0.65m//s`. |
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Answer» Differentiating `varphi(t)` with respect to time `(dvarphi)/(dt)=omega_z=2at` (1) For fixed axis rotation, the speed of the point A: `v=omegaR=2a tR` or `R=(v)/(2at)` (2) Differenciating with respect to time `w_t=(dv)/(dt)=2aR=v/t`, (using 1) But `w_t=v^2/R=(v^2)/(v//2at)=2atv` (using 2) So, `w=sqrt(w_t^2+w_n^2)=sqrt((v//t)^2+(2atv)^2)` `=v/tsqrt(1+4a^2t^4)` |
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