Saved Bookmarks
| 1. |
A simple pendulum of time period 1s and length l is hung from a fixed support at 0. Such that the bob is at a distance H vertically above A on the ground (figure) the amplitude is theta_(0) the string snaps at theta=(theta_0)/(2). Find the time taken by the bob to hit the ground. Also find distance from A where bob hits the ground. Assume theta_(0) to be small, so that sin theta_(0) approx theta_(0)" and"cos theta_(0) approx 1. . |
|
Answer» Solution :Two components of velocity Perpendicular component to the string `v_(y)= l omega sin ""(theta_0)/(2)` Prallel component to the string `v_(x)= l omega cos (theta_0)/(2)`  From `V= r omega` `therefore v= l omega` SUPPOSE at `t=0, THETA = theta_0` `therefore theta = theta_(0) cos omega t` but `T= 1s, therefore omega (2pi)/(T)= 2pi rad"/"s` `therefore theta= theta_(0) cos 2pi t"""........"(1)` Now at `t= t_(1)` time, `theta= (theta_0)/(2)` `therefore (theta_0)/(2)= theta_(0) cos 2pi t_(1)` `therefore (1)/(2) = cos 2pi t_(1)` `therefore 2pi t_(1) = (pi)/(3)` `therefore t_(1)= (1)/(6)s` Difference equation (1) with time, `(d theta)/(dt)= -theta_(0) 2pi sin 2pi t` `therefore t= t_(1)` time `omega = -2pi theta_(0) xx sin 2pi xx (1)/(6)` `=-2pi theta_(0) xx sin ""(pi)/(3)` `= -2pi theta_(0) xx sin"" (sqrt(3))/(3)` `therefore omega = -sqrt(3) pi theta_(0)` (-)ve shows that bob.s motion is towards, left. Since linear velocity is in the tangential direction. `therefore v= l omega""[therefore v= A omega]` `therefore v= -sqrt(3) pi theta_(0) l` The Y-component of linear velocity `v_(y)= -sqrt(3) pi theta_(0) l sin ""(theta_0)/(2)` and x-component of velocity `v_(x)= -sqrt(3) pi theta_(0) l cos"" (theta_0)/(2)` When string break, the height of bob from the ground, `H. = H+l(1- cos ""(theta_0)/(2))"""........"(2)` Time taken by the bob from this height (when string break) to ground it t, then `H.= v_(y)t+(1)/(2)G t^(2)` (all terms are (-)ve) `therefore (1)/(2) g t^(2) + sqrt(2) pi theta_(0) l sin ""(theta_0)/(2) l-H.=0` is a QUADRATIC formula of t `therefore triangle = b^(2) -4ac` `=3p^(2)theta_(0)^(2)sin^(2) ""(theta_0)/(2)+4xx(1)/(2)gxx H.` `=3pi^(2) theta_(0)^(2) l^(2) sin^(2) ""(theta_0)/(2) + 2gH.` but `theta_0` is small `therefore sin^(2) ""(theta_0)/(2)=0`, `triangle = 2gH.` `therefore sqrt(triangle) = sqrt(2gH.)` `therefore t= (-b pm sqrt(triangle))/(2a)` `= (-sqrt(3) pi theta_(0) l sin ""(theta_0)/(2) pm sqrt(2gH.))/(2xx (1)/(2)g)` but `sin""(theta_0)/(2)=0`, `t= (pm sqrt(2gH.))/(g)` `therefore t= sqrt((2H.)/(g))"""........"(3)` Now, in `H. = H+l(1-cos ""(theta_0)/(2)), theta_(0)` is small `cos ""(theta_2)/(2) approx 1` `therefore H.= H+l(1-1)` `therefore H.= H` `therefore t= sqrt((2H)/(g))""[" from equation (3) "]"""........."(4)` The bob covered distance in horizontal direction. `X= v_(x) t` `= -sqrt(3) pi theta_(0) l cos ""(theta_0)/(2)xx t` Since `theta_(0)` is small, `cos ""(theta_0)/(2) approx 1` and taking `t= sqrt((2H)/(g))`, `X= -sqrt(3) pi theta_(0) lxx sqrt((2H)/(g))` `therefore X= -sqrt((6H)/(g))xx pi theta_(0) l"""........"(5)` but string breaks, the bob is at RS distance from vertical at that time and `RS= l sin ""(theta_0)/(2)` `RS= (l theta_0)/(2)""[therefore ""sin (theta_0)/(2) approx (theta_0)/(2)]` When bob hit at ground, the distance from bob at that time `=(l theta_0)/(2)+X` `= (l theta_0)/(2) - sqrt((6H)/(g)) pi theta_(0) l` `= theta_(0) l((1)/(2)-pi sqrt((6H)/(g)))`. |
|