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?

Respuesta :

Answer with Explanation:

We are given that

[tex]Force,F=4.0 jN[/tex]

Mass of object,m=0.25 kg

Position vector,r=[tex](2.0i-2.0k) m[/tex]

Velocity vector,v=[tex](-5.0 i+5.0 k) m/s[/tex]

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

Angular momentum with respect to origin is given by

[tex]l=m(r\times v)[/tex]

Using the formula

[tex]l=0.25((2i-2k)\times (-5i+5k))[/tex]

[tex]l=0.25\times\begin{vmatrix}i&j&k\\2&0&-2\\-5&0&5\end{vmatrix}[/tex]

[tex]l=0.25(-j)(10-10)=0[/tex]

b.Torque acting on the object,[tex]\tau=r\times F[/tex]

[tex]\tau=(2i-2k)\times 4j=(8i\times j-8k\times j)=(8k+8i)Nm[/tex]

Where [tex]i\times j=k,k\times j=-i[/tex]

[tex]\tau=(8i+8k) Nm[/tex]

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