{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW8-phy2048-Spring-2014-solution

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}