Auniform thin rod of mass =3.47 kgm=3.47 kg pivots about an axis through its center and perpendicular to its length. two small bodies, each of mass m=0.263 kgm=0.263 kg , are attached to the ends of the rod. what must the length of the rod be so that the moment of inertia of the three-body system with respect to the described axis is =0.907 kg·m2i=0.907 kg·m2 ?

L= 1.468 m

Explanation:

The moment of inertia of the rod about its center is (1/12)m_RL^2

The moment of inertia of each of the two bodies about the described axes is m_B(L/2)2

Hence, the moment of inertia of three body system is (1/12)m_RL^2+ 2×m_B(L/2)^2 which is given to be equal to I_T

=> L^2[(1/12)m_R+m_B/2] = I_T

=> L2 = IT/(mR/12+ mB/2)

=> L = sqrt( 12I_T/(m_R+6m_B))

now putting the value of m_R= 3.47 kg

m_B= 0.263 kg

I_T = 0.97 kg/m^2

L= 1.468 m  is the length of the rod be so that the moment of inertia of the three-body system

dec 23, 2016 - a load of bricks weighing 680 n a distance of 10 m across a parking lot. if if the worker pushes the cart with a constant force of 209 n, what work= force x distance = 220 n x 10 m = 2200 n.m = 2200 j i don't know where the person got it from, but the initial part, work= force x distance me.