Calculation Question Answer on a new sheet of paper Identic

Calculation Question - Answer on a new sheet of paper Identical masses (m) are attached to opposite ends of a thin rod (length L and negligible mass). The masses are of equal and opposite charge(pmq). The system is located within a uniform electric field (E). The rod makes an angle (theta) relative to the direction of E and is relapsed from rest. a) what is the magnitude of the torque acting on the system? b) what is the maximum angular speed of the masses? c) Assume that the angle of displacement is small and derive the period of motion.

Solution

Charge on each end = q

Force on each charge F = Eq

Force on +q is to the right and on –q it is to the left

Arm of the couple = LSin()

Magnitude of the couple = EqLSin()

MI of the rod I = 2*m*(L/2)² = mL²/2

Work done by the couple when the rod rotated by an angle d

W = ò d

W = ò EqLSin()d = EqLCos()

this is equal to the rotational KE of the rod = I²/2, where is the angular speed.

(mL²/2)* ²/2 = EqLCos()

= sqrt(4Eq Cos()/mL)

it is maximum when =0

max angular speed = 2sqrt(Eq/mL)

Angular acceleration = /I = - EqLSin()/( mL²/2)

                                            = -2EqSin()/mL

Acceleration is proportional to the displacement for small dispalcement

= 2Eq/mL = -(2Eq/mL)

                             = -² ,   where =(2Eq/mL)

This a SHM motion with time period

T = 2/ = 2(mL/2Eq)