A rod of length I rotates with a small but uniform angular velocity co about its perpendicular bisector.

The right answer is (b) mω02a Explanation:  The frame of reference is rotating with a uniform angular speed ω0, so it has an acceleration towards the center and it is a non-inertial...

The correct answer is (a) T1 > T2. Explanation: Taking the rod of uniform mass per unit length = m . Mass beyond a distance r from the pivot = m(L-r). And this mass will be...

(d)1/2 mw2l. Explanation:  Since the rod rotates on the horizontal surface, the horizontal component of the force applied by the clamp is the centripetal force = mω²r = mω²(l/2) = ½mω²l.

(a) smallest at the top and gradually increases down the rod Explanation:  For a hanged heavy uniform rod, at any cross-section, the load is only of the part of the rod which...

t = 1 sec, r = 1 cm = 0.01 m  = 4 rd/s2 Therefore  ω= αt = 4 rad/s Therefore radial acceleration, An =ω2r = 0.16 m/s2 = 16 cm/s2

I1 = 6 kg-m2, ω1 = 2 rad/s , I2 = 5 kg-m2 Since external torque = 0 Therefore I1ω​1 = I2ω​2 ω2 = (6 × 2) / 5 = 2.4 rad/s

The correct answer is (a) zero.

(b) The rod and the experimenter move with the same speed v in opposite directions.

The correct answer is (a) T1 > T2. Explanation: Taking the rod of uniform mass per unit length = m . Mass beyond a distance r from the pivot = m(L-r). And this mass...

l = 20cm = 0.2m v = 10cm/s = 0.1m/s B = 0.10T (a) F = qvB = 1.6 × 10–19 × 1 × 10–1 × 1 × 10–1 = 1.6 × 10–21N (b)...