A massless rod of length 2l has equal point m | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The rate of change of angular momentum is equal to the net torque acting on the system,i.e.,$\left|\frac{dL}{dt}ight| = 	au_{	ext{net}}$.For e

Vedclass · Accepted Answer

$ml^2\omega^2\sin 2\alpha$

Vedclass · Answer

(B) The rate of change of angular momentum is equal to the net torque acting on the system,i.e.,$\left|\frac{dL}{dt}ight| = 	au_{	ext{net}}$.For each mass $m$ at a distance $l$ from the centre,the radius of the circular path is $r = l\sin\alpha$.The centripetal force acting on each mass is $F = mr\omega^2 = m(l\sin\alpha)\omega^2$.The torque $	au$ due to one mass about the centre is $	au = F 	imes r_{\perp}$,where $r_{\perp} = l\cos\alpha$ is the perpendicular distance from the axis of rotation to the force vector.Thus,$	au = (ml\sin\alpha\omega^2) 	imes (l\cos\alpha) = ml^2\omega^2\sin\alpha\cos\alpha$.Since there are two such masses on opposite sides of the centre,the torques produced by them are in the same direction.Therefore,the net torque is $	au_{	ext{net}} = 2	au = 2ml^2\omega^2\sin\alpha\cos\alpha$.Using the trigonometric identity $\sin 2\alpha = 2\sin\alpha\cos\alpha$,we get:$	au_{	ext{net}} = ml^2\omega^2\sin 2\alpha$.Hence,$\left|\frac{dL}{dt}ight| = ml^2\omega^2\sin 2\alpha$.

Similar Questions

$A$ torque of $50 \ N \cdot m$ causes a wheel to rotate from rest through $200 \ rad$ in $5 \ s$. The value of the angular acceleration produced is ...... $rad/s^2$.

$A$ rod of mass $M$ and length $L$ is placed in a horizontal plane with one end hinged about a vertical axis. $A$ horizontal force of $F = \frac{Mg}{2}$ is applied at a distance $\frac{5L}{6}$ from the hinged end. The angular acceleration of the rod will be:

What constant force must be applied tangentially at the equator to stop the Earth's rotation in one day?

$A$ torque of $30 \ Nm$ is applied for $15 \ s$ on a stationary wheel of mass $5 \ kg$. The moment of inertia of the wheel is $2 \ kg \ m^2$. The angular displacement of the wheel in $10 \ s$ is ....... $rad$.

$A$ uniform thin rod of length $L$ and mass $m$ is lying on a smooth horizontal table. $A$ horizontal impulse $P$ is suddenly applied perpendicular to the rod at one end. The total energy of the rod after the impulse is

Vedclass Test Series

Exam Paper Generator

Online Exam Module