Question

A swig sweat of mass M is connected to a fixed point P by a massless cord of length L. A child also of mass M sits on the seats and begins to swing with zero velocity at a position at which the cord makes a 60 angle with the verticle shown in figure 1. The swing continues down until the cord is exactly vertical at which time the child jumps off in a horizontal direction. The swing continues in the same direction until its cord makes a 45 angle with the vertical as shown in figure II: at that point it begins to swing in the reverse direction. With what velocity relative to the ground did the child leave the swing?

v-ar 지거 부+ 단시우+4bu+pg tose Fa 3. Go cos Ceo) 4F 45 ma ue FaureI Fa 1 A swing seat of mass M is connected to a fixed point P by a massless cord of length L. A child also of mass M sits on the seat and begins to swing with zero velocity at a position at which the cord makes a 60° angle with the vertical is shown in Figure I. The swing continues down until the cord is exactly vertical at which time the child jumps off in a horizontal direction. The swing continues in the same direction until its cord makes a 45° angle with the vertical as shown in Figure II: at that point it begins to swing in the reverse direction. With what velocity relative to the ground did the child leave the swing? (cos 45 = sin 45° 2/2, sin 30° cos 60° 1/2, cos 30° sin 60° = 3/2) A1# F-I a Tr

Answer 1

When seat+child is at initial position, energy of system is  

E1 = PE = (M+Mg L - L cos 60°)

( since the seat+ child is at a height of )

11 = (L - L cos 60°)

E1 = PE = 2MgL(1 – cos 60°

When the swing in vertical, energy of system is

En = F(M + M)u? = Mus

Conserving the energy of system at two positions,  

2MgL(1 – cos 60°) = M02

v= 2gL(1 - cos 60°)

Now the child jumps off the seat. His velocity after jumping off the seat is $v_1$ . And velocity of seat is $v_2$ ,

The seat starting with velocity $v_2$ goes to a height

h2 = L(1 - cos 45°)

Conserving the energy of seat at vertical position and at height $h_2$ ,

Mvz = MgL(1 – cos 45°)

U2 = V2gL(1 - cos 45º)

Conserving the momentum of child and seat before and after the child jumps off the seat,

Mυ + Mυ = Μυη + Mus

Vi = 20 – uz

V1 = 212gL(1 - cos 60°) - 2gL(1 - cos 45º)

-=2V77 - 1991 ( 19 )

(74–)764 – 16/7 = 1

Speed of child relative to ground when he leaves the swing is

(74–)764 – 16/7 = 1