JEE Main20203 Sep 2020Evening ShiftPhysicsRotational MotionActual
A uniform rod of length ' ℓ ' is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft rotates at angular speed ω the rod makes an angle θ with it (see figure). To find θ equate the rate of change of angular momentum (direction going into the paper) m ℓ 2 12 ω 2 sin θ about the centre of mass (CM) to the torque provided by the horizontal and v
Options
- Acos θ = 2 g 3 l ω 2
- Bcos θ = g 2 ℓ ω 2
- Cc o s θ = g ℓ ω 2
- Dcos θ = 3 g 2 ℓ ω 2
Correct answer
D. cos θ = 3 g 2 ℓ ω 2
Step-by-step solution
Torque of centrifugal force τ cf = dm . x   sin   θω 2 xcos   ϑ = m l ω 2   sin   θcos   θ ∫ 0 l x 2 dx τ ef = m l 2 ω 2 sinθ   cosθ 3 τ mg = τ cf mg . l 2 sin   θ = m l 2 ω 2   sinθ   cos   θ 3 cos   θ = 3 g 2 l ω 2