5 kg m2 s at t 20 s 11p51 no external torques

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Unformatted text preview: out of the page, or along the +z axis, perpendicular to the plane of Fig. 11- 41. And for the 6.5 kg particle, we find l 2 = r⊥ 2 mv2 = (1.5 m ) ( 6.5 kg ) ( 2.2 m/s ) = 21.4 kg ⋅ m2 s. !! And we use the right- hand rule again, finding that this ( r2 × p2 ) is into the page, or in the –z direction. (a) The two angular momentum vectors are in opposite directions, so their vector sum is the difference of their magnitudes: L = ℓ1 − ℓ2 = 9.8 kg ⋅ m2 s . (b) The direction of the net angular momentum is along the +z axis. 11P35. (a) We note that r r dr = 8.0t ^ – (2.0 + 12t)^ i j v= dt with SI units understood. From Eq. 11- 18 (for the angular momentum) and Eq. 3- ^ 30, we find the particle’s angular momentum is 8t2k . Using Eq. 11- 23 (relating its → ^ time- derivative to the (single) torque) then yields τ = (48t k) N ⋅ m . (b) From our (intermediate) result in part (a), we see the...
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