A particle of mass 2, kg is moving such that | vedclass

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

Question

Angular momentum $L = m$ $(v 	imes r)$

$\begin{array}{l}
 = 2kg\,\left( {\frac{{dr}}{{dt}} 	imes r} ight) = 2\,kg\left( {4t\,j 	imes 5i

Vedclass · Accepted Answer

$ - 80\hat k$

Vedclass · Answer

Angular momentum $L = m$ $(v 	imes r)$

$\begin{array}{l}
 = 2kg\,\left( {\frac{{dr}}{{dt}} 	imes r} ight) = 2\,kg\left( {4t\,j 	imes 5i - 2{t^2}\hat j} ight)\
 = 2\,kg\,\left( { - 20\,t\,\hat k} ight) = 2\,kg 	imes  - 20 	imes 2\,{m^{ - 2}}{s^{ - 1}}\hat k\
 =  - 80\,\hat k
\end{array}$

A particle of mass $2\, kg$ is moving such that at time $t$, its position, in meter, is given by $\overrightarrow r \left( t \right) = 5\hat i - 2{t^2}\hat j$ . The angular momentum of the particle at $t\, = 2\, s$ about the origin in $kg\, m^{-2}\, s^{-1}$ is

Similar Questions

Obtain $\tau = I\alpha $ from angular momentum of rigid body.

The position of a particle is given by : $\overrightarrow {r\,} = (\hat i + 2\hat j - \hat k)$ and momentum $\overrightarrow P = (3\hat i + 4\hat j - 2\hat k)$. The angular momentum is perpendicular to

A solid sphere rolls without slipping on a rough surface and the centre of mass has a constant speed $v_0$. If the mass of the sphere is $m$ and its radius is $R$, then find the angular momentum of the sphere about the point of contact

$A$ 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 moves with a constant velocity in $X-Y$ plane. Its possible angular momentum w.r.t. origin is