By Gregory R.D.

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**Extra resources for Solutions manual for Classical mechanics**

**Example text**

The initial condition r D 0 when t D 0 gives C D 0 so that ur ut ; D sin 1 a Ua that is, Ua ut rD sin : u a This is the solution for r as a function of t . Daniel will be caught when r D a, that is, when u ut D : sin a U If U u, this equation has the real solution tD a sin u 1 u U c Cambridge University Press, 2006 42 Chapter 2 Velocity, acceleration and scalar angular velocity and so Daniel will be caught after this time. Since D ut =a, the polar equation of the path of the lion is rD Ua sin : u In order to recognise this equation as a circle, we express it in Cartesian coordinates.

And the speed of the centre C of the link at time t . Solution Let be the angle between the rod and the negative x-axis. 14) is ! D . a sin / . D D xP a sin b cos t b cos t D 1=2 a sin a2 a2 cos2 b cos t D a2 2 2 b sin t 1=2 : This is the angular velocity of the rod at time t . X; Y /. Then X D 12 a cos ; Y D 21 a sin ; and so XP D YP D Hence XP 2 C YP 2 D 1 4 1 a sin 2 1 a cos 2 P ; P : a2 sin2 P 2 C D 41 a2 P 2 D 1 4 a2 cos2 P 2 2 a2 b 2 cos2 t : 4 a2 b 2 sin2 t c Cambridge University Press, 2006 Chapter 2 Velocity, acceleration and scalar angular velocity The speed of C at time t is therefore abj cos t j 1=2 : 2 a2 b 2 sin2 t c Cambridge University Press, 2006 54 55 Chapter 2 Velocity, acceleration and scalar angular velocity Problem 2 .

Daniel will be caught when r D a, that is, when u ut D : sin a U If U u, this equation has the real solution tD a sin u 1 u U c Cambridge University Press, 2006 42 Chapter 2 Velocity, acceleration and scalar angular velocity and so Daniel will be caught after this time. Since D ut =a, the polar equation of the path of the lion is rD Ua sin : u In order to recognise this equation as a circle, we express it in Cartesian coordinates. This is made easier if both sides are multiplied by r .