As we know that fulcrum is shifted to one side by 1.33 cm
so here let say left side mass is m1 while on other side mass is m2
so we will have
now in order to find the percentage we have
so it is 10.5 % less mass
Both papers will reach the ground at the same time. CORRECT ANSWER
Answer:
Explanation:
The problem is related to rotational motion . So we shall find out rotational kinetic energy .
K E = 1/2 x I ω²
ω is the final angular velocity
Moment of inertial of the disk
I ₁ = 1/2 m r²
= .5 x 165 x 2.93²
= 708.25 kgm²
Moment of inertial of the person
I₂ = mr²
= 62.5 x 2.93²
= 536.55 kgm²
ω₂ = v / R
= 3.11 / 2.93 rad /s
At the time of jumping , law of conservation of angular momentum will apply
I₁ ω₁ + I₂ω₂ = (I₁ + I₂)ω
708.25 x0.691 + 536.55 x ( 3.11 / 2.93 ) = ( 708.25 + 536.55 ) ω
ω = 0 .85 rad/ s
K E = 1/2 x I ω²
= .5 x ( 708.25 + 536.55 ) ( .85 )²
449.68 J
