# How does centripetal force changes the direction of velocity isn't it a violation to Newton's second law of motion?

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

• Yup, you're correct – Jim Apr 12 '17 at 19:21
• So basically we are ignoring the small change in speed at all times!! – Matt Apr 12 '17 at 19:22
• 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 – Jim Apr 12 '17 at 19:28
• Can you please elaborate a little. – Matt Apr 12 '17 at 19:31
• I really thought I had. What part was confusing? – 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.