(10-21) I = Mx² + m(L-x)² à (dI/dx) = 2Mx - 2m(L-x) = 0. Solving this
for x gives à x = mL /(M+m). Now (d² I / dx² ) = 2m+2M >0 so this value
for x is a minimum. Now if you take x = mL /(M+m) [which is the location
of the CoM] and insert it into I = Mx² + m(L-x)² you get à
I = M[mL/(m+M)]² + m[L - mL/(m+M)] ²= [mM /(m+M)] L² = m L²
Where m = mM/(m+M)

(10-26) Torque eqn. à å (t ) = 0 = mg(3r) -Tr.
Force eqn. à å F =0 = 2T - M g sin 45° à T = (Mg sin 45° )/ 2
T = (1500 *9.8 * sin 45° )/2 = 5197 N
Now the torque eqn. Yields m = T/3g = 5197/(3*9.8) = 177 kg

(10-33) (a) Find the velocity of the CoM using à (KE + U)_i = (KE +U)_f à
0+mgR = (1/2) I w ² +0 à w =Ö (2mgR/I) = Ö (2mgR/[3m R²/2]) = Ö (4g/3R)
now v_cm= Rw = R Ö (4g/3R) =2Ö (Rg/3)
(b) v_low = 2 v_cm = 4Ö (Rg/3)
(c ) Everything in part (a) will stay the same except now I=2mR² (by
the parallel-axis theorem) so w = Ö (g/R) so v_cm= Rw = R Ö (g/R) =Ö (Rg)

(10-43) (a) Use energy conservation (KE + U)_i = (KE +U)_f à
0+ (1/2)mgL = (1/2) I w ² + 0 à w =Ö (mgL/I) =Ö (mgL/[mL²/3])
w =Ö (3g/L)
(b) t = Ia so that in the horizontal (mgL/2) = (m L² /3) a à a =3g/2L
(c ) a_x = a_cent = (L/2) w ² =-(3g/2) ; a_y = a_t = (L/2) a = 3g /4
(d) Using F=ma we have R_x = m a_x = -(3mg /2)
R_y - mg = -m a_y à R_y = mg / 4

(10-48) (a) KE_f = (1/2) M (v_f)² + (1/2) I (w _f)² ; U_f =Mgh_f = 0
KE_i = (1/2) M (v_i)² + (1/2) I₀ (w _i)² = 0 ; U_i =Mgh_i
So W_nc = E_f - E_i à -fd = KE_f+U_f - KE_i -U_i = (1/2) M (v_f)² + (1/2) I (w _f)² - Mgh_i
Now using f=m N = m M g cos q ; w =(v/r) ; h_i = d sinq ; I = (1/2) m r²
in the above gives
-m M g cosq = (1/2)M v² + (mr² /2)(v/r)² /2 - M g d sinq à
(1/2)[M + (m/2)] v² = M g d sinq - m M g cosq à
v² = 2M g d [(sinq - m cosq ) /{(m/2) + M}]
v = Ö [4g d {M/(m+2M)} (sinq - m cosq )]
(b) v² = (v₀)² +2a d à v² = 2a d à a = v² /2d
a = 2g d {M/(m+2M)} (sinq - m cosq )

(10-53) (a) From the given conditions 3.5 = w ₀ e^-0 so w ₀ =3.5 and
2 = 3.5 e^{-s (9.3)} so s = (-1/9.3)*ln(2/3.5) = 0.0602 sec^-1 Now
a = d w /dt = (d /dt) [3.5 e^-0.0602*t ] = -3.5*0.0602* e^-0.0602*3.0 = -0.176 rad/sec²
(b) q = ò ₀^(t=2.5) w ₀ e^{-s t}dt = (3.5 / -0.0602) [e^-0.0602*2.5 -1] = 8.12 rad = 1.29 rev
(c ) In part (b) just let t=2.5 à t=¥ so q = (3.5 / -0.0602) [e^-¥ -1]
q = (3.5 / -0.0602) [0 -1] = 58.14 rad = 9.25 rev

(10-55) From the forces acting on m we can find ma = F_total = T -mg where T is the
string tension. So T = ma-mg. Now since m undergoes uniform acceleration we
have y = (1/2) a t² or a = -(2y / t²) à T = m(-[2y/t²] + g). Now the spool will experience
two torques -- one from the tension, T, and one from the friction between the spool and
the rod. So Ia =TR - t _f à t _f =TR - Ia . Now TR= mR( g - [2y/t²])
and a =a/R =(2y / Rt²) so t _f = mR(g - [2y/t²]) -(2y I/ Rt²)
Finally I = (1/2) M[R² + (R/2)² ] = (5/8) M R² . Using this we
Finally get t _f = R[m(g - [2y/t²]) -(5My / 4t²) ]

(11-5A) KE = (1/2) M v² + (1/2) I w ². Now w = v/R and I = (2/5) MR²
so KE = (1/2) M v² + (1/2) (2/5) MR² (v/R)² = (1/2) M v² +(1/5) Mv²
KE = (7/10) M v²

(11-7) (a) A´ B = -17 k (using eqn. 11.14 with the given vectors)
(b) | A´ B| =| A| | B| sin q à 17 = 5Ö (13) sinq à sinq =17 / 5Ö (13) à
q = sin^-1 (17 / 5Ö (13) ) = 70.5°

(11-18) Looking at the forces on m in the horizontal direction gives
T sinq = m v² / r ; Looking at the forces in the vertical direction gives
T cosq = m g Taking the ratio of these two equations gives
T sinq / T cosq = sinq / cosq = [m v² / r] /m g = v² / r g
Solving for v gives v = Ö [r g (sinq / cosq )]
Now L = m v r sin 90° = m r Ö [r g (sinq / cosq )]
L = Ö [m² r³ g (sinq / cosq )].
Finally r=l sinq à L =Ö [m² l³ g (sin⁴q/ cosq )].