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In a Khan Academy video about centripetal force calculations: here, the presenter explains

If a force doesn't point radially inwards or outwards you don't include it in the calculation

I perfectly get that for the case given when gravity has no component in the radial direction, but what if the weight force DID have a component radially?

enter image description here

Sorry for the crappy image [this is a vertical circle], but you can see that this weight force can be decomposed to have a radial component. So question is, do we include this part in calculations? Additionally, shouldn't the centripetal force be the net force acting on an object?

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    $\begingroup$ I haven't viewed the video, but based on what you present I would say that a better phrase would be "If a force doesn't have a component pointing radially ..." $\endgroup$ – garyp Jan 28 '19 at 1:52
  • $\begingroup$ In considering the radial component of the force balance, you would include the component of the gravitational force acting in the radial direction. $\endgroup$ – Chet Miller Jan 28 '19 at 2:53
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You can resolve the forces in to radial (toward and away from the center) and tangential (along the circle at that instant) parts.

In the diagram, those are the dashed lines. You reason about the two parts separately. The radial part is about the circular nature of the motion.

The tangential part acts to speed up or slow down the circular motion, changing the V along the circle. If that tangential part is zero, the speed remains constant. In the case in the figure, it will be speeding up.

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  • $\begingroup$ so in a calculation would you take the component radially outward and subtract it from the normal force to get the "net" centripetal force inward? $\endgroup$ – John Hon Jan 28 '19 at 1:39
  • $\begingroup$ If you know N, yes. But often you only know that the motion is constrained to a circle, so you subtract to find N. $\endgroup$ – Bob Jacobsen Jan 28 '19 at 1:43
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The acceleration has two components. One that is radial $a_r$ and one that is centripetal $a_c$. Presuming that the object is bound to the circular path, each of these acceleration components is related to the tangential velocity $v_T$ in their own way.

$$a_c = \frac{v^2_T}{r}$$ $$a_T = \frac{dv_T}{dt}$$

($dv_T /dt$ means "instantaneous rate of change of $v_T$. If you haven't seen this notation before, you can think of it like $\Delta v_T / \Delta t$ when $\Delta t$ is taken to be very small). Essentially, any non-centripetal component of net force acting on an object contributes to it speeding up or slowing down while it travels along the circular path.

You can also write similar expressions for the net force components by multiplying by the mass of the object $m$.

$$F_{\text{net, }c} = m\frac{v^2_T}{r}$$ $$F_{\text{net, }T} = m\frac{dv_T}{dt}$$

In cases when motion is known to have constant speed or is known to have no tangential acceleration, disregarding these components of force doesn't do harm since we know the forces will cancel.

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The centripetal force is only a function of the tangential velocity. A force that is acting tangentially will accelerate or decelerate you and cause your angular/tangential velocity to increase or decrease, but it is only the velocity at any moment in time that is related to the centripetal force and not how much the velocity is increasing or decreasing.

So it is true that any force not acting radially is irrelevant to centripetal force.

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