Giancoli 5th Ed.
Hwk 10 (Due 4/26/23) 10 point hwk
Ch. 11 P. 1,2,3,5,36,42,48
MisConQ
Problems:
1. L = Iw = MR^2w = 3.41 kg m^2/s
2. a) L = Iw = 1/2 MR^2 w = 7.1 kg m^2/s
b) torque = Delta L / Delta t = (0 - 7.125)/(6.0-0)
t = -1.2 Nm
3. L_i = L_f so Iw + I(0) = (I+I)w_f
(I/2I) w = w_f or w_f = 1/2 w
5. L_i = L_f so I_i w_i = I_f w_f
and I_f = I_i (w_i)/w_f) = (4.9)(0.66/2.5) = 1.3 kg m^2
36. a) L = r_perp p = dmv
or p = mvd directed into the plane of the page.
b) L = rxp = 0 or r_perp = 0. So L = 0
42. a) Can find L_y (component of L along y-axis) using
L_y = I_1 w_1 + I_2 w_2
where I_1 = I_2 = M(.24m)^2 and w_1=w_2=4.5 rad/s
this gives L_y = 0.249 kg m^2/s
OR, you can find the total L = L_x i + L_y j using
L_tot = r_1 x p_1 + r_2 x p_2
with r_1 x p_1 = (-.24i+.21j+0)x(0i+0j-[.48*1.08]k)
and r_2 x p_2 = ( .24i-.21j+0)x(0i+0j+[.48*1.08]k)
so L_tot = 0.218i + 0.249 j kg m^2/sec
and the L_y is just the j-component, L_y = 0.249 kg m^2/sec
b) At the moment shown in the figure, the L_tot vector
points up and to the right. (It will precess with time such
that the L_y component stays fixed while L_x,L_z change.)
The angle between L_tot and the y-axis is
theta = tan^-1 (L_x,max/L_y) = tan^-1 (.218/.249)
= 41 degrees
48. L_i = L_f so L_bull = (L_stick + L_bull)_final
= m_bul v_i (l/4) = I_stick w + m_bul v_f (l/4)
Solve for omega, w = m_bul (v_i-v_f)(l/4) / I_stick
use I_stick = (1/12) Ml^2
then w = 4.3 rad/sec
Please check
Prob. 5 (1pt), 48 (1pt) Tot= 10 pts
MisConQ. --- (0.5 pt each)