Saved Bookmarks
| 1. |
A body rotates about a stationary axis. If the angular deceleration is proportional to square root of angular speed, then the mean angularspeed of the body, givenomega_0 as the initial angular speed, is |
|
Answer» `(omega_0)/(sqrt2)` angular deceleration `prop sqrt("angular SPEED")` i.e.,` (- d omega)/(DT) prop sqrt (omega) ` or`(-d omega )/(dt) = K sqrt(omega )` where, k is constant or` - int_(omega_0)^(omega) (d omega)/(sqrt (omega)) = k int_(0)^(t) dt ` `- [2 sqrt(omega)]_(omega_0)^(omega) = k [t]_(0)^(t) ` `rArr - 2[ sqrt(omega) - sqrt(omega_0) ] =kt ` `sqrt(omega) - sqrt(omega_0) =- (kt)/(2)` `rArrsqrt(omega) = sqrt(omega_0) - (kt)/(2) "" ...(i)` When then the total TIME of rotation, ` tau = t = (2 sqrt( omega_0) )/(k)` ` therefore ` Mean angular speed, `< baromega > = (int omega dt )/(int dt) = ( int_(0)^(2 sqrt(omega_0) // k) ( omega_0 + (k^2 t^2)/(4) - kt sqrt(omega_0) dt ))/(2 sqrt(omega_0) // k )` ` = ( [ omega_0 t + (k^2 t^3)/(12) - k/2 sqrt(omega_0) t^2 ]_(0)^(2sqrt(omega_0) // k) )/(2 sqrt(omega_0) // k ) ` After SOLVING , we get ` = (omega_0)/(3)` |
|