A long thin uniform rod of length 1.50 m is to be suspended from a frictionless pivot located at some point along the rod so that its pendulum motion takes 3.00 s. How far from the center of the rod should the pivot be located?

0.087 m

Explanation:

Length of the rod, L = 1.5 m

Let the mass of the rod is m and d is the distance between the pivot point and the centre of mass.

time period, T = 3  s

the formula for the time period of the pendulum is given by

   .... (1)

where, I is the moment of inertia of the rod about the pivot point and g is the acceleration due to gravity.

Moment of inertia of the rod about the centre of mass, Ic = mL²/12

By using the parallel axis theorem, the moment of inertia of the rod about the pivot is

I = Ic + md²

Substituting the values in equation (1)

12d² -26.84 d + 2.25 =  0

d = 2.15 m , 0.087 m

d cannot be more than L/2, so the value of d is 0.087 m.

Thus, the distance between the pivot and the centre of mass of the rod is 0.087 m.

a

explanation:

because they are cartillidge

0.11 m/s and 0.28 m/s should be correct