A 3.0 kg particle with velocity v = (5.0 m/s) i -(6.0 m/s) j is at x = 3.0 m, y

Question

A 3.0 kg particle with velocity v = (5.0 m/s) i -(6.0 m/s) j is at x = 3.0 m, y = 8.0 m. It ispulled by a 7.0 N force in the negative x direction.

(a) What is the angular momentum of the particle about theorigin?

_________kg m^2/s k



(b) What torque about the origin acts on the particle?

_________Nm k



(c) At what rate is the angular momentum of the particlechanging?

_____kg m^2/s^2



(a) What is the angular momentum of the particle about theorigin?

_________kg m^2/s k



(b) What torque about the origin acts on the particle?

_________Nm k



(c) At what rate is the angular momentum of the particlechanging?

_____kg m^2/s^2

Explanation / Answer

Mass m = 3.0 kg
velocity v = (5.0 m/s) i - (6.0 m/s) j position r = 3 i + 8 j Force F = -7.0 i (a) the angular momentum of the particle about the origin= m ( r X v ) rX v = ( 3i+8j ) X ( 5 i -6 j )         = ( 3 * -6 ) ( iX j) + ( 8 * 5 ) ( j X i )         = -18 k -40 k         = -58 k So, angular mementum L = m ( rXv ) = -174 k kg m^2/sk (b) torque about the origin acts on the particle T = r XF                                                                           =(3i+8j) X(-7i)                                                                          = ( 8 * -7 ) ( jXi)                                                                          = 56 k N (c) rate is the angular momentum of the particlechanging = value of torque                                                                                       = 56 kg m^2/s^2

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