A thin uniform rod is pivoted at O. It rotate | vedclass

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

Question

(B) The torque $	au$ required to keep the angular velocity $\omega$ constant is given by the rate of change of angular momentum $L$. $L = I\omega$. S

Vedclass · Accepted Answer

Vedclass · Answer

(B) The torque $	au$ required to keep the angular velocity $\omega$ constant is given by the rate of change of angular momentum $L$. $L = I\omega$. Since $\omega$ is constant,$	au = \frac{dL}{dt} = \omega \frac{dI}{dt}$. The moment of inertia $I$ of the system (rod + insect) at time $t$ is $I = I_{	ext{rod}} + I_{	ext{insect}}$. $I_{	ext{insect}} = m r^2$,where $r = vt$ is the distance of the insect from $O$ at time $t$. So,$I(t) = I_{	ext{rod}} + m(vt)^2 = I_{	ext{rod}} + mv^2t^2$. Differentiating with respect to $t$,we get $\frac{dI}{dt} = 2mv^2t$. Therefore,the torque is $	au = \omega (2mv^2t) = (2m\omega v^2)t$. This shows that $	au$ is directly proportional to $t$ for $0 \le t \le T$. At $t = T$,the insect stops,so $\frac{dI}{dt} = 0$,and the torque becomes zero. Thus,the graph should be a straight line passing through the origin for $0 \le t \le T$ and then drop to zero. This corresponds to the shape shown in graph $B$.

Similar Questions

$A$ same torque is applied to a disc and a ring of equal mass and radii. Then:

$A$ hollow sphere of mass $1 \,kg$ and radius $10 \,cm$ is free to rotate about its diameter. If a force of $30 \,N$ is applied tangentially to it,its angular acceleration is (in $rad/s^2$):

Vedclass Test Series

Exam Paper Generator

Online Exam Module