Newton's second law is a vector law. When when we resolve it in component form along the x, y and z axes we can conclude that force changes only the component of velocity along it, for example if the only force is along x axis then only the velocity along x changes but not other two. So why, in uniform circular motion, does the centripetal force changes the direction of velocity even though it is perpendicular to velocity?

My reasoning is that at any instant say at $t = 0$ the force is along radius and perpendicular to velocity, at $t = dt$ the velocity perpendicular to force is unchanged both in magnitude and direction but a new velocity is gained $dv$ in $dt$ time which is along the radius and now the resultant velocity has the same magnitude as before approximately but a different direction.

Is my reasoning correct or is there some a other explanation?

  • $\begingroup$ Yup, you're correct $\endgroup$ – Jim Apr 12 '17 at 19:21
  • $\begingroup$ So basically we are ignoring the small change in speed at all times!! $\endgroup$ – Matt Apr 12 '17 at 19:22
  • $\begingroup$ nope, the magnitude remains constant. The amount $dv$ is so small that it's actually zero. But after each $dt$, there's a corresponding $d\theta$ change in the radial direction. Over the next $dt$, there's a bit more added to the velocity in one direction but subtracted from the velocity in the other direction $\endgroup$ – Jim Apr 12 '17 at 19:28
  • $\begingroup$ Can you please elaborate a little. $\endgroup$ – Matt Apr 12 '17 at 19:31
  • $\begingroup$ I really thought I had. What part was confusing? $\endgroup$ – Jim Apr 12 '17 at 19:33

One can always express the position of some object with respect to some origin as a vector in as $\mathbf r = r \hat r$ where $\mathbf r$ is the object's position vector, $r$ is the magnitude of that position vector, and $\hat r$ is the unit vector parallel to $\mathbf r$. Differentiating with respect to time yields $\mathbf v \equiv \dot{\mathbf r} = \dot r \hat r + r \dot{\hat r}$.

If the radial distance is constant, $\dot r$ is identically zero. But what about $\dot{\hat r}$? This is a unit vector, which is a special case of a constant length vector. Consider a vector $\mathbf x$ whose length is constant with respect to time: $||\mathbf x||^2 = \mathbf x \cdot \mathbf x= \text{const}$. Differentiating with respect to time yields $\mathbf x \cdot \dot{\mathbf x} = 0$. In other words, the time derivative of a constant length vector is either zero or is normal to that constant length vector. A unit vector is obviously a special case of a constant length vector.


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