A particle, mass 0.25 kg is at a position () m, has a velocity () m/s, and is subject to a force () N. What is the magnitude of

Question

baherus [9]

2 years ago

12

A particle, mass 0.25 kg is at a position () m, has a velocity () m/s, and is subject to a force () N. What is the magnitude of

the torque on the particle about the origin

Physics

1 answer:

NeX [460] · Answer 1 · 2022-08-19T19:50:42+03:00

<h2>Question </h2>

A particle, mass 0.25 kg is at a position (<em>-7i + 7j + 5k</em>) m, has a velocity (<em>6i - j + 4k</em>) m/s, and is subject to a force (<em>-5i + 0j - k</em>) N. What is the magnitude of the torque on the particle about the origin?

<h2>Answer:</h2>

47.94Nm

<h2>Explanation:</h2>

The torque (τ) on a particle subject to a force (represented as force vector F) at a position (represented as position vector r) about the origin is given by the cross product of the position vector r for the point of application of a force and the force F. i.e

τ = r x F

Given:

r = (-7i + 7j + 5k) m

F = (-5i + 0j - k) N

| i j k |

r x F = | -7 7 5 |

| -5 0 -1 |

r x F = i(-7 - 0) - j(7+25) + k(0+35)

r x F = i(-7) - j(32) + k(35)

r x F = -7i - 32j + 35k

Therefore the torque τ = -7i - 32j + 35k

The magnitude of the torque is therefore;