ANSWERS. 233: 



Pages 96, 97. 



(i) The integration gives tan \ (ir + 0) = tan J (TT +0 ) * * ' , 

 which gives t = <x> for = IT. 



(5) The particle will not leave the circle if N remains positive; 



^=oif 0)50! = ---- 



3 I 



(6) Greater than 17-94 ft. per sec. 



(8) The tangential force is mg sin 0, if 6 is the inclination of the 

 tangent to the horizon ; as sin = dz/ds, we have 



dv dz j d^s 



-*=**' or *^=**' 



whence \ i? ^ v$ = g(z z ) . The result also follows from the equa- 

 tion of kinetic energy, since the force- function is C7= mgz -f C. 



(9) The equation of motion is the same as (8), Art. 175, except. 

 that g is replaced by g sin a. 



(10) The particle performs oscillations of period 2 TT//U, if F= (Jr. 



Page 98. 



Taking 6 for q, we find Q = (mg + pnO) K ; hence the inte- 



gration of (17) gives tf = h 2 KgO !-0 2 . On the other hand, (14) 



m 



dt 



m 



Pages 102, 103. 



(3) O) 9'8 in.; (b) 1-23 oz. 



(4) As x andjj' are small of the first order, z differs from r, by (5), 

 only by a small quantity of the second order ; hence z = o, z/r i, so 

 that the third of the equations (6) gives N=mg] hence x = gx/r, 

 y = gy/r. Integrating and putting "*fgjr /x, we find x = C^ sin /x/ 

 4- C 2 cos //,/, y = D l sin //,/ + D% cos /x/. Solving for sin and cos, squar- 

 ing and adding, we find an ellipse as the required path, just as in 

 Art. 121. 



