m = mass of the circular hoop
r = radius of the hoop
I = moment of inertia of the hoop
moment of inertia of the hoop about the center of hoop is given as
I = m r²
k = distance of the point of suspension from center of mass = r
using parallel axis theorem
I' = moment of inertia of hoop about the point of suspension
I' = I + m k²
I' = m r² + m k²
I' = m r² + m r²
I' = 2 m r²
Time period of oscillation for the hoop is given as
T = 2π sqrt(I'/mgk)
T = 2π sqrt(2 m r²/(mgr))
T = 2π sqrt(2 r/g)
since 2r = diameter = d
T = 2π sqrt(d/g)
Answer:
Hey guess what there's no question genius
Explanation:
As the first astronaut throws the ball, lets assume it goes with v velocity and the mass of the ball be m
the momentum comes out be mv, thus to conserve that momentum the astronaut will move opposite to the direction of the ball's motion with the velocity mv/M (where M is the mass of the astronaut).
