Question:
Suppose two children are using a uniform seesaw that is 3.00 m long and pivoted at 2 m from the left edge, and a 30.0 kg child sits 1.40 m to the right of the the center of mass of the seesaw. Seesaw weighs 5 kg.
a. Calculate where the second 18.0 kg child must sit to balance the seesaw. [ 5 points]
b. How far will the child be from the center of mass of the seesaw? [ 2 points]
c. How far will the child be from the left edge of the seesaw? [3 points]
Answer:
(a)
Length of seesaw l = 3 m
Pivot point is at 2m from left
Centre of mass of seesaw is at 1.5 m, therefore, distance of centre of mass from pivot point is
lc = 2 - 1.5 m =0.5m
The child sits 1.40 m to the right of the the center of mass of the seesaw. So, distance of child from pivot point is
lR = 1.40 - 0.5 m = 0.9 m, towards right
mass of child at right mR = 30kg
mass of seesaw,mc= 5 kg
mass of child at left, mL = 18 kg
distance of left child from pivot point = lL
Child on the left and centre of mass of seesaw provide a anticlockwise torque whereas student on the right provide a clockwise torque.
For rotational symmetry
(Ans)
The child should sit at 1.36 m from the pivot point, towards left
(b)
The center of mass of the seesaw is at 0.5m from the pivot
Distance of child to the centre of mass is
(Ans)
(c)
The distance of left edge to the pivot point is 2m
Distance of child from the left edge is
(Ans)