Torque of centrifugal force τcf=dm.xsinθω2xcosϑ=lmω2sinθcosθ∫0lx2dx
τef=3ml2ω2sinθcosθ
τmg=τcf
mg.2lsinθ=3ml2ω2sinθcosθ
cosθ=2lω23g

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 θwith it (see figure). To find θ equate the rate of change of angular momentum (direction going into the paper) 12mℓ2ω2sinθ about the centre of mass (CM) to the torque provided by the horizontal and vertical forces FHand Fv about the CM. The value of θ is then such that:
Held on 3 Sept 2020 · Verified 6 Jul 2026.
cosθ=3lω22g
cosθ=2ℓω2g
cosθ=ℓω2g
cosθ=2ℓω23g
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
Net gravitational force at the center of a square is found to be $F_{1}$ when four particles having mass $M, 2 M, 3 M$ and $4 M$ are placed at the four corners of the square as shown in figure and it is $F_{2}$ when the positions of $3 M$ and $4 M$ are interchanged. The ratio $\frac{F_{1}}{F_{2}}$ is $\frac{\alpha}{\sqrt{5}}$. The value of $\alpha$ is $\_\_\_\_$. 
A particle of mass 2 kg is projected vertically upward with a speed of 30 m/s. The maximum height reached by the particle is (g = 10 m/s²):
Two projectiles are projected with the same initial velocities at the $15^\circ$ and $30^\circ$ with respect to the horizontal. The ratio of their ranges is $1:x$. The value of $x$ is:
In an experiment the values of two spring constants were measured as $k_{1}=(10 \pm 0.2) \mathrm{N} / \mathrm{m}$ and $k_{2}=(20 \pm 0.3) \mathrm{N} / \mathrm{m}$. If these springs are connected in parallel, then the percentage error in equivalent spring constant is :
The surface tension of a soap solution is $3.5 \times 10^{-2}$ N/m. The work required to increase the radius of a soap bubble from $1$ cm to $2$ cm is $\alpha \times 10^{-6}$ J. The value of $\alpha$ is _____. ($\pi = 22/7$)
Work through every JEE Main Mechanics PYQ, year by year.