Question

part 1 A thin, hollow sphere of radius r = 0.480 m and mass m = 13.5 kg turns counterclockwise about a vertical axis through its center (when viewed from above), at an angular speed of 2.90 rad/s. What is its vector angular momentum about this axis? (Enter the magnitude in kg. m2/s.) magnitude kg . m/s direction ---Select--- part 2 A particle of mass 0.500 kg is attached to the 100-cm mark of a meterstick of mass 0.200 kg. The meterstick rotates on the surface of a frictionless, horizontal table with an angular speed of 6.00 rad/s. (a) Calculate the angular momentum of the system when the stick is pivoted about an axis perpendicular to the table through the 50.0-cm mark. kg : m?/s (b) Calculate the angular momentum of the system when the stick is pivoted about an axis perpendicular to the table through the 0-cm mark. kg m/s

Answer 1

Moment of inertia about centre axis of sphere is Ios 2 mr² angular speed w = 2.90 rad's I. C angular momentum lo- L. 2 mo?w x 13-565) (8-48m) (2.9086ty) Lo - Tlo-6-01 kgm'ls direction : out of the page when wiewed from above 2 mass of stick = M = 0.2ooky length of rod. L=1m mass of particle=m= 0.5ookg angular speed 6.0ored /s و کیا moment of inestia about socm mask is I, I, I = ML² tmd? 2 | 2 m dr Jocm = 0.5m → 7,- 0.2006) (em)? +6.500kg) (0.5m)? 12 I, 0:141667 Kgos? angular momentum 4=I, w Lic (0.141667 kgm?) (Grades) allis 0.85 kgm/s © moment of inestia about ocm mark is I & Iz น I = M1² 2 MU 3 tha2 here dolm I 2 (0.2ks) (im? +(0.56g) (m)? - 0.5133 tym? 3 angular momentum L2= I w (0.5133 kam") (Grades) اد + L₂= 3.08 kgm/ Scanned with CamScanner