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 →

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?

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Tryphena 5 years 2021-08-17T07:48:42+00:00 1 Answers 127 views 0

Answers ( )

  1. haidang
    0
    2021-08-17T07:50:10+00:00
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    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

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