angular momentum (urgent)

57) M.I. of combined rod + particle = 0.82 M/12 + (M/3) x 0.42 = 0.32M/3

So total angular momentum = 0.32M/3 x 20 = 6.4M/3

At the moment when the particle leaves the rod:

Velocity of particle = vp, giving an angular momentum of 0.4Mvp/3

Velocity of the end of the rod = vp - 6, goving an angular momentum of (0.64M/12) (vp - 6)/0.4 = (2M/15) (vp - 6)

Hence:

(2M/15) (vp - 6) + 0.4Mvp/3 = 6.4M/3

2vp - 12 + 2vp = 32

4vp = 44

vp = 11 m/s

61) Total angular momentum = 0.12 x 2.4 = 0.288 kg m2/s

Total M.I. of rod + putty = 0.12 + 0.2 x 0.62 = 0.192

Hence new angular speed = 0.288/0.192 = 1.5 rad/s

65) Total M.I. of merry + boy = 150 + 30 x 22 = 270 kg m2

Angular momentum contributed by the ball = 1 x 2 cos 37 x 12 = 19.2 kg m2/s.

Combined M.I. of merry + boy + ball = 270 + 1 x 22 = 272 kg m2.

So angular speed = 19.2/272 = 0.070 rad/s

67a) Total M.I. = 2 x 2 x 0.252 = 0.25 kg m2.

Angular momentum contributed by the wad = 0.05 x 0.25 x 3 = 0.0375 kg m2/s

New total M.I. = 0.25 + 0.05 x 0.252 = 0.253125 kg m2.

So angular speed = 0.0375/0.253125 = 0.148 rad/s

b) K.E. of wad before collision = 0.5 x 0.05 x 32 = 0.225 J

K.E. after collision = 0.5 x 0.253125 x 0.1482 = 0.00278 J

So the ratio is 0.00278/0.225 = 0.0123

c) Suppose that it will turn through θ before mom stops again, then:

G.P.E. gain by the ball w/o wad = 2 x 0.25 sin θ = 0.5 sin θ

G.P.E. loss by the ball with wad = 2.05 x 0.25 sin θ = 0.5125 sin θ

So when it mom stops, K.E. = 0, i.e.

G.P.E. gain by the ball w/o wad - G.P.E. loss by the ball with wad = 0.00278 J which is the initial K.E. just after collision.

0.5 sin θ - 0.5215 sin θ = 0.00278

-0.0215 sin θ = 0.00278

sin θ = - 0.129

θ = 181