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abhijitlohiya
Jul18-06, 02:17 AM
A particle in UCM goes from a point A to B. The change in it's momentum is 2j and change in the centripetal force is 8i .Where i and j are unit vectors along X and Y axis. What is the angular velocity of the particle ?

andrevdh
Jul18-06, 08:44 AM
The answer is 4 rad/s.

andrevdh
Jul19-06, 02:28 AM
The magnitude of the momentum and force vectors do not change as the vectors rotate. The first diagram shows the two momentum vectors and their change as the particle rotated through an angle \phi (I chose clockwise rotation. Mirror the vectors in the y-axis for anticlockwise rotation). The vectors form an isosceles triangle. The second diagram shows the two momentum vectors of the particle as it rotated from point A to point B.

andrevdh
Jul20-06, 02:54 AM
Note that the force vectors form a similar triangle with the same angle \phi between them. This is because the vectors are rotating together. From here on it is only two little more steps to the answer, so I am reluctant to provide any more information - the answer should be quite obvious from my two posts.

andrevdh
Jul22-06, 08:06 AM
Thought I should post the answer for comlpeteness sake.

The two triangles formed by the momentum and force vectors are similar since all three angles are the same. This means that the ratio of their sides (the magnitude of the vectors) - are equal. Therefore:

\frac{mv^2}{8r} = \frac{mv}{2}

so that

\frac{v}{r} = 4

which is the required angular velocity of the particle.