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A round cone with half-angle alpha=30^@ and the radius of the base R=5.0cm rolls uniformly and without slipping over a horizontal plane as shown in fiugre. The cone apex is hinged at the point O which is one the same level with the point C, the cone base centre. The velocity of point C is v=10.0cm//s. Find the moduli of (a) the vector of the angular velocity of the cone and the angle it forms with the vertical, (b) the vector of the angular acceleration of the cone. |
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Answer» Solution :(a) Let the axis of the cone (OC) rotates in anticlockwise sense with constant angular VELOCITY `vecomega` and the cone itself about it's own axis (OC) in clockwise sense with angular velocity `vecomega_0`(figure). Then the resultant angular velocity of the cone. `vecomega=vecomega+vecomega_0` (1) As the rolling is pure the magnitudes of the vectors `vecomega` and `vecomega_0` can be easily found from figure. `omega'=(v)/(Rcot alpha)`, `omega_0=v//R` (2) As `vecomega_|_vecomega_0` from Eq. (1) and (2) `omega=sqrt(omega^('^2)+omega_0^2)` `sqrt(((v)/(Rcotalpha))^2+(v/R)^2)=(v)/(Rcosalpha)=2*3rad//s` (B) Vector of angular acceleration `vecbeta=(dvecomega)/(dt)=(d(vecomega+vecomega_0))/(dt)`(as `vecomega=` constant.) The vector `vecomega_0` which rotates about the `OO^'` axis with the angular velocity `vecomega`, retains i magnitude. This increment in the time interval dt is equal to `|dvecomega_0|=omega_0*omega^'dt` or in vector from `dvecomega_0=(vecomegaxxvecomega_0)dt`. Thus `vecbeta=vecomegaxxvecomega_0` The magnitude of the vector `vecbeta` is equal to `beta=omega^'omega_0` (as `vecomega_|_vecomega_0`) So, `beta=(v)/(Rcotalpha)(v)/(R)=v^2/R^2 tan alpha=2*3rad//s` 
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