A disc of radius $5\, m$ is rotating with angular frequency $10\, rad / sec .$ A block of mass $2\, kg$ to be put on the disc friction coefficient between disc and block is $\mu_{ k }=0.4,$ then find the maximum distance from axis where the block can be placed without sliding (in $cm$)

  • [AIIMS 2019]
  • A

    $2$

  • B

    $3$

  • C

    $4$

  • D

    $6$

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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 $\omega$ the rod makes an angle $\theta$ with it (see figure). To find $\theta$ equate the rate of change of angular momentum (direction going into the paper ) $\frac{ m \ell^{2}}{12} \omega^{2} \sin \theta \cos \theta$ about the centre of mass $(CM)$ to the torque provided by the horizontal and vertical forces $F_{H}$ and $F_{V}$ about the CM. The value of $\theta$ is then such that:

  • [JEE MAIN 2020]

A uniform rod of mass $15\,kg$ and length $5\,m$ is held stationary with the help of a light string as shown. Then tension in the string is ......... $N.$

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A rod of mass $1\ kg$ , length $1\ m$ is suspended horizontally with the help of two ideal strings as shown in figure. First a mass is suspended to the left most end keeping rod horizontal, than a second mass is suspended to the right most end again keeping horizontal orientation. The maximum total mass that can be suspended in that way keeping horizontal orientation of rod ....... $kg.$

A frame of reference that is accelerated with respect to an inertial frame of reference is called a non-inertial frame of reference. A coordinate system fixed on a circular disc rotating about a fixed axis with a constant angular velocity $\omega$ is an example of a non-intertial fiame of reference. The relationship between the force $\vec{F}_{\text {rot }}$ experienced by a particle of nass in moving on the rotating disc and the force $\vec{F}_{\text {in }}$ experienced by the particle in an inertial frame of reference is

$\vec{F}_{\text {rot }}=\vec{F}_{\text {in }}+2 m\left(\vec{v}_{\text {rot }} \times \vec{\omega}\right)+m(\vec{\omega} \times \vec{r}) \times \vec{\omega},$

where $\vec{v}_{\text {rot }}$ is the velocity of the particle in the rotating frame of reference and $\bar{r}$ is the position vector of the particle with respect to the centre of the disc.

Now consider a smooth slot along a diameter of a disc of radius $R$ rotating counter-clockwise with a constant angular speed $\omega$ about its vertical axis through its center. We assign a coordinate system with the origin at the center of the disc, the $x$-axis along the slot, the $y$-axis perpendicular to the slot and the $z$-axis along the rotation axis $(\vec{\omega}=\omega \hat{k})$. A sm a $1$ block of mass $m$ is gently placed in the slot at $\vec{r}=(R / 2) \hat{i}$ at $t=0$ and is constrained to move only along the slot.

(Image)

($1$) The distance $r$ of the block at time $t$ is

($A$) $\frac{R}{4}\left(e^{\omega t}+e^{-\omega t}\right)$   ($B$) $\frac{R}{2} \cos \omega t$   ($C$) $\frac{R}{4}\left(e^{2 \omega t}+e^{-2 \omega t}\right)$

($D$) $\frac{F}{2} \cos 2 \omega t$

($2$) The net reaction of the disc on the block is

($A$) $\frac{1}{2} m \omega^2 R\left(e^{2 \omega t}-e^{-2 \omega t}\right) \hat{j}+m g \hat{k}$

($B$) $\frac{1}{2} m \omega^2 R\left(e^{\omega t}-e^{-a t t}\right) j+m g k$

($C$) $-m \omega^2 R \cos \omega t \hat{j}-m g \hat{k}$

($D$) $m \omega^2 R \sin \omega t \hat{j}-m g \hat{k}$

Give the answer quetioin ($1$) ($2$)

  • [IIT 2016]