Question #75613
Description:
A uniform rod of mass M and length L is lying on a horizontal frictionless surface if a horizontal impulse I perpendicular to length of the road is applied at one end of the road the other end begins to move with speed V then find the magnitude of the impulse
Solution.
draw a diagram of the problem
r c r_c r c
r c x = L 2 cos α , r c y = L 2 sin α \mathrm {r} _ {\mathrm {c x}} = \frac {\mathrm {L}}{2} \cos \alpha , \quad \mathrm {r} _ {\mathrm {c y}} = \frac {\mathrm {L}}{2} \sin \alpha r cx = 2 L cos α , r cy = 2 L sin α
find the velocity components as derived from the radius of the vector
d r c x d t = v c x = − L 2 sin α d α d t , d r c y d t = v c y = L 2 cos α d α d t \frac {\mathrm {d} \mathbf {r} _ {\mathrm {c x}}}{\mathrm {d t}} = \mathbf {v} _ {\mathrm {c x}} = - \frac {\mathrm {L}}{2} \sin \alpha \frac {\mathrm {d} \alpha}{\mathrm {d t}}, \frac {\mathrm {d} \mathbf {r} _ {\mathrm {c y}}}{\mathrm {d t}} = \mathbf {v} _ {\mathrm {c y}} = \frac {\mathrm {L}}{2} \cos \alpha \frac {\mathrm {d} \alpha}{\mathrm {d t}} dt d r cx = v cx = − 2 L sin α dt d α , dt d r cy = v cy = 2 L cos α dt d α
horizontally, the lower end moves at a speed v = c o n s t \mathbf{v} = \mathrm{const} v = const
L cos α = L − v t \mathrm {L} \cos \alpha = \mathrm {L} - \mathrm {v t} L cos α = L − vt
find the derivative (2)
d ( L cos α ) d x = − L sin α d α d t = − v = 2 v c x ( 2 ) , = > v c x = − v 2 , = > sin α = v L d α d t \frac {\mathrm {d} (\mathrm {L} \cos \alpha)}{\mathrm {d} x} = - \mathrm {L} \sin \alpha \frac {\mathrm {d} \alpha}{\mathrm {d} t} = - v = 2 v _ {\mathrm {c x}} (2), = > v _ {\mathrm {c x}} = - \frac {v}{2}, = > \sin \alpha = \frac {v}{\mathrm {L} \frac {\mathrm {d} \alpha}{\mathrm {d} t}} d x d ( L cos α ) = − L sin α d t d α = − v = 2 v cx ( 2 ) , => v cx = − 2 v , => sin α = L d t d α v
it is obvious that the vertical speed as an opposite to the horizontal speed
v c y = − v c x cot α v _ {c y} = - v _ {c x} \cot \alpha v cy = − v c x cot α
from equation (1) we find the cotangent
c o t a n α = c o s 2 α 1 − c o s 2 α = ( L − v t L ) 2 1 − ( L − v t L ) 2 = L − v t L 2 − L 2 + 2 L v t − v 2 t 2 = L − v t 2 L v t − v 2 t 2 c o t a n \alpha = \sqrt {\frac {c o s ^ {2} \alpha}{1 - c o s ^ {2} \alpha}} = \sqrt {\frac {(\frac {L - v t}{L}) ^ {2}}{1 - (\frac {L - v t}{L}) ^ {2}}} = \frac {L - v t}{\sqrt {L ^ {2} - L ^ {2} + 2 L v t - v ^ {2} t ^ {2}}} = \frac {L - v t}{\sqrt {2 L v t - v ^ {2} t ^ {2}}} co t an α = 1 − co s 2 α co s 2 α = 1 − ( L L − v t ) 2 ( L L − v t ) 2 = L 2 − L 2 + 2 Lv t − v 2 t 2 L − v t = 2 Lv t − v 2 t 2 L − v t v c y = v c x cot α = v 2 L − v t 2 L v t − v 2 t 2 v _ {c y} = v _ {c x} \cot \alpha = \frac {v}{2} \frac {L - v t}{\sqrt {2 L v t - v ^ {2} t ^ {2}}} v cy = v c x cot α = 2 v 2 Lv t − v 2 t 2 L − v t
then the momentum of the rod is a vector with coordinates
M v c → = < v c x , v c y > = < − M v 2 , M v 2 L − v t 2 L v t − v 2 t 2 > \overrightarrow{\mathrm{Mv}_c} = < v_{cx}, v_{cy} > = < -M\frac{v}{2}, M\frac{v}{2} \frac{L - vt}{\sqrt{2Lvt - v^2t^2}} > Mv c =< v c x , v cy >=< − M 2 v , M 2 v 2 Lv t − v 2 t 2 L − v t > ∣ M v c ∣ → = M v c x 2 + v c y 2 = M L v 2 1 2 L v t − v 2 t 2 \overrightarrow{|Mv_c|} = M \sqrt{v^2_{cx} + v^2_{cy}} = ML \frac{v}{2} \frac{1}{\sqrt{2Lvt - v^2t^2}} ∣ M v c ∣ = M v c x 2 + v cy 2 = M L 2 v 2 Lv t − v 2 t 2 1
Answer
M v c → = < v c x , v c y > = < − M v 2 , M v 2 L − v t 2 L v t − v 2 t 2 > \overrightarrow{\mathrm{Mv}_c} = < v_{cx}, v_{cy} > = < -M\frac{v}{2}, M\frac{v}{2} \frac{L - vt}{\sqrt{2Lvt - v^2t^2}} > Mv c =< v c x , v cy >=< − M 2 v , M 2 v 2 Lv t − v 2 t 2 L − v t > ∣ M v c ∣ → = v c x 2 + v c y 2 = M L v 2 1 2 L v t − v 2 t 2 \overrightarrow{|Mv_c|} = \sqrt{v^2_{cx} + v^2_{cy}} = ML \frac{v}{2} \frac{1}{\sqrt{2Lvt - v^2t^2}} ∣ M v c ∣ = v c x 2 + v cy 2 = M L 2 v 2 Lv t − v 2 t 2 1
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