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edugen.wileyplus.com the figure he. WileyPLUS how to take Q Search Halliday, Fundamentals of Physics, 10e PHYS PRI OF NANO SCIENCE 1 (NENGI Assignment Gradebook ORION Downloadable eTextbook ent MESSAGE HY INSTRUCTOR FULL SCREEN PRINTER VERSION BACK NEXT Chapter 11, Problem 033 = (7.42 mis li-12.98 m/s is at x . 6.95 m, y_ 4.75 m. it is pulled by a the negative x direction. About the origin, what are (a) the particle's angular momentum, (b) the torque A 1.30 kg particle with velocity acting on the particle, and (c) the rate at which the angular momentum is changing? (a) Number k Units (b) Number Units (c) Number Units ly LINK TO SAMPLE PROBLEM LINK TO SAMPLE PROBLEM

Explanation / Answer

Given that
m = 1.3 kg

v = (7.42 m/s) i - (2.98 m/s) j

r = 6.95 i + 4.75 j

F = -3.69 i

(a)
the angular momentum of the particle about the origin = m*( r * v )

r*v = ( 6.95 i + 4.75 j ) *( 7.42 i - 2.98 j )

= ( 6.95 * - 2.98 ) ( i* j) + ( 4.75 * 7.42 ) ( j X i )

= - 20.711 k - 35.245 k

= - 55.95 k

So,
angular mementum L = m*( r*v )

= - 72.735 k kg m^2/sk

(b)
torque about the origin acts on the particle

T = r * F

= (6.95 i + 4.75 j) *(- 3.69 i)

= ( 4.75 * - 3.69 ) ( j*i)

= 17.53 k N

(c)
rate is the angular momentum of the particle changing = value of torque

= 17.53 kg m^2/s^2

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