A disc of mass $M$ and radius $R$ is rolling with angular speed $\omega $ on a horizontal plane as shown. The magnitude of angular momentum of the disc about the origin $O$ is
A disc of mass $M$ and radius $R$ is rolling with angular speed $\omega $ on a horizontal plane as shown. The magnitude of angular momentum of the disc about the origin $O$ is
- A
$\frac {1}{2} MR^2\omega $
- B
$MR^2\omega $
- C
$\frac {3}{2} MR^2\omega $
- D
$2MR^2\omega $
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The position vectors of radius are $2\hat i + \hat j + \hat k$ and $2\hat i - 3\hat j + \hat k$ while those of linear momentum are $2\hat i + 3\hat j - \hat k.$ Then the angular momentum is
A particle is moving along a straight line parallel to $x$-axis with constant velocity. Find angular momentum about the origin in vector form
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The potential energy of a particle of mass $m$ at a distance $r$ from a fixed point $O$ is given by $\mathrm{V}(\mathrm{r})=\mathrm{kr}^2 / 2$, where $\mathrm{k}$ is a positive constant of appropriate dimensions. This particle is moving in a circular orbit of radius $\mathrm{R}$ about the point $\mathrm{O}$. If $\mathrm{v}$ is the speed of the particle and $\mathrm{L}$ is the magnitude of its angular momentum about $\mathrm{O}$, which of the following statements is (are) true?
$(A)$ $v=\sqrt{\frac{k}{2 m}} R$
$(B)$ $v=\sqrt{\frac{k}{m}} R$
$(C)$ $\mathrm{L}=\sqrt{\mathrm{mk}} \mathrm{R}^2$
$(D)$ $\mathrm{L}=\sqrt{\frac{\mathrm{mk}}{2}} \mathrm{R}^2$
The potential energy of a particle of mass $m$ at a distance $r$ from a fixed point $O$ is given by $\mathrm{V}(\mathrm{r})=\mathrm{kr}^2 / 2$, where $\mathrm{k}$ is a positive constant of appropriate dimensions. This particle is moving in a circular orbit of radius $\mathrm{R}$ about the point $\mathrm{O}$. If $\mathrm{v}$ is the speed of the particle and $\mathrm{L}$ is the magnitude of its angular momentum about $\mathrm{O}$, which of the following statements is (are) true?
$(A)$ $v=\sqrt{\frac{k}{2 m}} R$
$(B)$ $v=\sqrt{\frac{k}{m}} R$
$(C)$ $\mathrm{L}=\sqrt{\mathrm{mk}} \mathrm{R}^2$
$(D)$ $\mathrm{L}=\sqrt{\frac{\mathrm{mk}}{2}} \mathrm{R}^2$
- [IIT 2018]
$A$ time varying force $F = 2t$ is applied on a spool rolling as shown in figure. The angular momentum of the spool at time $t$ about bottommost point is:
$A$ time varying force $F = 2t$ is applied on a spool rolling as shown in figure. The angular momentum of the spool at time $t$ about bottommost point is:
The angular momentum of a particle performing uniform circular motion is $L$. If the kinetic energy of partical is doubled and frequency is halved, then angular momentum becomes
The angular momentum of a particle performing uniform circular motion is $L$. If the kinetic energy of partical is doubled and frequency is halved, then angular momentum becomes