234 



ANSWERS. 



Page 112. 



(1) The differential equation is r <oV + " sin to/ = o. If f=r Q> 

 dr/dt = v Q at the time /= o, we find 



( g\* ( \ -<* 8 



(2) Let be the angle, at the time /, between the radius CP t drawn 

 from the centre C of the circle to the particle P, and the diameter OCA 

 through the fixed point O. Then, taking O as origin, and the initial 

 position OC$AQ of the diameter OCA (when P is at A ) as axis of x, 

 we find 



d z O . _. . fdO^ 



. + o> 2 sin = o, whence = 2 o> 2 cos + C. 

 <// \"*/ 



As the absolute initial velocity is zero, we find = 2 <>. Hence, 

 <9 2 = 4 o> 2 cos 2 1 (9, and finally, sin 1 ^ = ^ ~ g _^, or tan J (TT + ^) = ^>' . 



(3) Let x z +y* =a 2 (i + /") 2 be the equation of the circle, whence 

 -\- y8y = o ; the equation of motion is xBx -f j'S) ; = o ; eliminating 



#, S>>, and integrating, we find xy yx av^. The equation of the 

 circle gives xx +y' = a z a(i + at) ; hence, 



a(i + a/) 2 * = ^(i + a/)^ - swV(i +0 2 -^ 2 - 

 'To integrate, put i + a/= r, and then put x = r. The result is 



x = a(i+at) cos ^ - , y= a(i + at) sin 



' * 



Pages 134, 135. 



The square of the radius of inertia is : 



(1) K- 



(2) OH/ 2 ; (b)\tf; W T V/ 2 ; W 

 (3)i/^ (8) \a 



(4) iV 2 - (9) i 



(5) 2\ 2 - (10) i^ 



(6) \h\ (n) ^^ 



(7) / 2 - (12) 



