Honors Solutions to Sheet 3A

Rotational Problem Sheet 3 A solutions

1. f = ½ so mega = 2 pi x ½ = pi rad/s; vtan = omega r = 2 pi m/s;
ac = omega vtan = 2 (pi)sqrd

2. Mac = Mmoon omega sqrd r = F gravity = G Mp Mmoon/ rsqrd so

omega =sqrt (GMp/rcubed) = 5.77E-6 rad/s; T = 2pi/omega

3. a) at bottom lowest PE greatest KE; b) at bottom greatest vtan so greatest ac and also Fg is away from center and ac is up so FT = Mac + Mg

4. Mac = M vtan sqrd /r = Qvtan B so r = Mv/QB = 3.34E-27*3E6/(1.6E-19*5)

M ac = M omega vtan r = QvtanB so omega = QB/M = 1.6E-19*5/3.34E-27

5. a) omega = vtan/r = 1.5 rad /s b) ac = vtan sqrd/r = 18 m/s2 c) 65x18 =1170N up d) Fseat= Mac – Fg =1170 – - Mg = 1820 N up