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A particle of mass m moves in a certain plane P due to a force F whose magnitude is constant and whose vector rotates in that plane with a constant angular velocity omega. Assuming the particle to be stationary at the moment t=0, find: (a) its vecocity as a function of time, (b) the distance covered by the particle between two successive stops, and the mean velocity over this time. |
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Answer» Solution :Let US FIX x-y co-ordinate system to the given plane, taking x-axis in the direction along which the force vector was oriented at the moment `t=0`, then the fundamental equation of DYNAMICS expressed via the projection on x and y-axes gives, `Fcos omegat=m(dv_x)/(dt)` (1) and `Fsin omegat=m(dv_y)/(dt)` (2) (a) Using the condition `v(0)=0`, we obtain `v_x=(F)/(momega)sin omegat` (3) and `v_y=(F)/(momega)(1-cosomegat)` (4) Hence, `v=sqrt(v_x^2+v_y^2)=((2F)/(momega))|sin ((omegat)/(2))|` (b) It is seen from this that the velocity `v` turns into zero after the time interval `Deltat`, which can be found from the relation, `omega(Deltat)/(2)=pi`. Consequently, the sought distance, is `s=underset(0)overset(Deltat)intvdt=(8F)/(momega^2)` Average velocity, `lt v ge(intvdt)/(intdt)` So, `lt v geunderset(0)overset(2pi//omega)int (2F)/(momega)sin ((omegat)/(2))dt//(2piomega)=(4F)/(pimomega)` 
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