A circular platform rotates in a horizontal p | vedclass

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

Question

(C) The angular momentum $L$ of the system (platform + tortoise) about the axis of rotation is conserved because there is no external torque acting on

Vedclass · Accepted Answer

Vedclass · Answer

(C) The angular momentum $L$ of the system (platform + tortoise) about the axis of rotation is conserved because there is no external torque acting on the system.$L = I(t) \omega(t) = 	ext{constant}$Here, $I(t)$ is the moment of inertia of the system at time $t$.Let $R$ be the radius of the platform, $M$ its mass, and $m$ the mass of the tortoise.The moment of inertia of the platform is $I_p = \frac{1}{2}MR^2$.The tortoise moves along a chord. Let the distance of the chord from the center be $d$. At time $t$, the distance of the tortoise from the center is $r(t) = \sqrt{d^2 + (vt)^2}$, where $v$ is the constant speed of the tortoise relative to the platform.The total moment of inertia is $I(t) = I_p + mr(t)^2 = I_p + m(d^2 + v^2t^2)$.Since $L = I(t) \omega(t)$, we have $\omega(t) = \frac{L}{I_p + m(d^2 + v^2t^2)}$.As the tortoise moves from the edge towards the center (or vice versa), the distance $r(t)$ first decreases until it reaches the point closest to the center (at $t=0$ if the chord passes through the center, or at some $t$ if it doesn't) and then increases.When $r(t)$ is minimum, $I(t)$ is minimum, and therefore $\omega(t)$ is maximum.Conversely, when $r(t)$ is maximum (at the edges), $I(t)$ is maximum, and $\omega(t)$ is minimum.Since the tortoise starts at the edge, $r(t)$ decreases as it moves towards the center, causing $I(t)$ to decrease and $\omega(t)$ to increase. After passing the point closest to the center, $r(t)$ increases, causing $I(t)$ to increase and $\omega(t)$ to decrease.Thus, the angular velocity $\omega(t)$ increases and then decreases.

Similar Questions

$A$ rigid spherical body is spinning around an axis without any external torque. Due to a change in temperature,the volume increases by $1\%$. Its angular speed:

$A$ ring of mass $M$ and radius $R$ is rotating about its axis with angular velocity $\omega$. Two identical bodies each of mass $m$ are now gently attached at the two ends of a diameter of the ring. Because of this,the kinetic energy loss will be

If the radius of a rotating disc is suddenly halved while its mass remains the same,its angular velocity will become:

Vedclass Test Series

Exam Paper Generator

Online Exam Module