Unformatted text preview: ′Fz − z ′Fy ˆ + ( z ′Fx − x ′Fz ) ˆ + x ′Fy − y′Fx k. i
j ( ) ( ) !!
(a) Here, r ′ = r . Dropping the primes in the above expression, we set (with SI units understood) x = 0, y = 0.5, z = –2.0, Fx = 2.0, Fy = 0, and Fz = –3.0. Then we obtain rrr
ˆ
ˆ
ˆ
τ = r × F = −1.5i − 4.0j −1.0k N ⋅ m. ( ) ( ) ! !!
r
ˆ
ˆ
(b) Now r ′ = r − ro where ro = 2.0i − 3.0k. Therefore, in the above expression, we set x′ = −2.0, y′ = 0.5, z′ = 1.0, Fx = 2.0, Fy = 0 , and Fz = −3.0. Thus, we obtain r r r ˆ
τ = r ′ × F = −1.5 ˆ − 4.0 ˆ − 1.0k N ⋅ m. i
j 3 PHY2048 Spring 2014 HW8 11P29. For the 3.1 kg particle, Eq. 11 21 yields l 1 = r⊥1mv1 = ( 2.8 m ) (3.1 kg ) (3.6 m/s ) = 31.2 kg ⋅ m2 s. !!
Using the right hand rule for vector products, we find this ( r1 × p1 ) is...
View
Full Document
 Spring '14
 Physics, Kinetic Energy, Mass, Work, Eq.

Click to edit the document details