A small particle of mass $m$ is projected at an angle $\theta $ with the $x-$ axis with an initial velocity $v_0$ in the $x-y$ plane as shown in the figure. At a time $t < \frac{{{v_0}\,\sin \,\theta }}{g}$, the angular momentum of the particle is
$ - mg\,{v_0}{t^2}\,\cos \,\theta\, \hat j$
$ mg\,{v_0}t\,\cos \,\theta\, \hat k$
$ - \frac{1}{2}\,mg\,{v_0}{t^2}\,\cos \,\theta \,\hat k$
$\frac{1}{2}\,mg\,{v_0}{t^2}\,\cos \,\theta \,\hat i$
$A$ particle of mass $0.5\, kg$ is rotating in a circular path of radius $2m$ and centrepetal force on it is $9$ Newtons. Its angular momentum (in $J·sec$) is:
A particle of mass $M=0.2 kg$ is initially at rest in the $x y$-plane at a point $( x =-l, y =-h)$, where $l=10 m$ and $h=1 m$. The particle is accelerated at time $t =0$ with a constant acceleration $a =10 m / s ^2$ along the positive $x$-direction. Its angular momentum and torque with respect to the origin, in SI units, are represented by $\vec{L}$ and $\vec{\tau}$, respectively. $\hat{i}, \hat{j}$ and $\hat{k}$ are unit vectors along the positive $x , y$ and $z$-directions, respectively. If $\hat{k}=\hat{i} \times \hat{j}$ then which of the following statement($s$) is(are) correct?
$(A)$ The particle arrives at the point $(x=l, y=-h)$ at time $t =2 s$.
$(B)$ $\vec{\tau}=2 \hat{ k }$ when the particle passes through the point $(x=l, y=-h)$
$(C)$ $\overrightarrow{ L }=4 \hat{ k }$ when the particle passes through the point $(x=l, y=-h)$
$(D)$ $\vec{\tau}=\hat{ k }$ when the particle passes through the point $(x=0, y=-h)$
A particle of mass $2\, kg$ is on a smooth horizontal table and moves in a circular path of radius $0.6\, m$. The height of the table from the ground is $0.8\, m$. If the angular speed of the particle is $12\, rad\, s^{-1}$, the magnitude of its angular momentum about a point on the ground right under the centre of the circle is ........ $kg\, m^2\,s^{-1}$
A solid sphere of mass $500\,g$ and radius $5\,cm$ is rotated about one of its diameter with angular speed of $10\,rad \, s ^{-1}$. If the moment of inertia of the sphere about its tangent is $x \times 10^{-2}$ times its angular momentum about the diameter. Then the value of $x$ will be ..............
A ball of mass $1 \,kg$ is projected with a velocity of $20 \sqrt{2}\,m / s$ from the origin of an $x y$ co-ordinate axis system at an angle $45^{\circ}$ with $x$-axis (horizontal). The angular momentum [In $SI$ units] of the ball about the point of projection after $2 \,s$ of projection is [take $g=10 \,m / s ^2$ ] ( $y$-axis is taken as vertical)