Question

At one instant, force → F = 4.0 ˆ j N acts on a 0.25 kg object that has position vector → r = ( 2.0 ˆ i − 2.0 ˆ k ) m and velocity vector → v = ( − 5.0 ˆ i + 5.0 ˆ k ) m / s . About the origin and in unit-vector notation, what are (a) the object’s angular momentum and (b) the torque acting on the object?

Answer 1

Answer with Explanation:

We are given that

$Force,F=4.0 jN$

Mass of object,m=0.25 kg

Position vector,r=$(2.0i-2.0k) m$

Velocity vector,v=$(-5.0 i+5.0 k) m/s$

a.We have to find the angular momentum of object.

Angular momentum with respect to origin is given by

$l=m(r\times v)$

Using the formula

$l=0.25((2i-2k)\times (-5i+5k))$

$l=0.25\times\begin{vmatrix}i&j&k\\2&0&-2\\-5&0&5\end{vmatrix}$

$l=0.25(-j)(10-10)=0$

b.Torque acting on the object,$\tau=r\times F$

$\tau=(2i-2k)\times 4j=(8i\times j-8k\times j)=(8k+8i)Nm$

Where $i\times j=k,k\times j=-i$

$\tau=(8i+8k) Nm$