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During non-uniform circular motion, the direction of net acceleration is not in the direction of the centripetal acceleration, then why does a particle still move in a circular path, please explain.

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Think of a car going round a circular track. At an instant when it has speed $v$ it has an acceleration of magnitude $\frac{v^2}{r}$ towards the centre of the circle. The car is gaining velocity towards the circle centre. But suppose that, at this instant, the driver is making the car go faster. The car will also be gaining velocity in a direction tangential to the circle. That doesn't interfere with its gaining of velocity towards the circle centre.

Looking at it in terms of forces, the road is exerting a frictional force on the powered wheels of the car that has both a forward component, giving the car its increase in speed, and a sideways component towards the circle centre, allowing the car to go at speed $v$ in a circle of radius $r$.

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  • $\begingroup$ As the speed increases, the component of the friction force toward he center of the curve must also increase. $\endgroup$
    – R.W. Bird
    Commented Sep 30, 2020 at 13:29
  • $\begingroup$ That's right, it will increase. My answer referred to a particular instant; I wasn't trying to give an account of what happened over time. $\endgroup$ Commented Sep 30, 2020 at 13:32
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Let's look at a position vector in general polar coordinates $$\vec r(t)=r(t)\pmatrix{\cos\theta(t)\\\sin\theta(t)}$$ Now define \begin{align} \hat r&=\pmatrix{\cos\theta\\\sin\theta}\\ \hat \theta&=\pmatrix{-\sin\theta\\\cos\theta} \end{align} to make our lives easier. I dropped the time dependence but remember it's still there. You can check that \begin{align}\dot{\hat r}&=\hat\theta \dot\theta\\\dot{\hat \theta}&=-\hat r \dot\theta\end{align} where a dot indicates a time derivative. A general acceleration then becomes \begin{align}\ddot{\vec r}(t)=&\hat r(\ddot r-r\dot\theta^2)+\\&\hat\theta(r\ddot\theta+2\dot r\dot \theta).\end{align} Again this is a nice exercise to proof. For circular motion the radius is constant so $\dot r=0$. What's left of the general acceleration is \begin{align}\ddot{\vec r}(t)=&\hat r(-r\dot\theta^2)+\\&\hat\theta(r\ddot\theta).\end{align} The $-r\dot\theta^2$ term is the usual centripetal acceleration which points towards the center. You might recognise it if you plug in $\dot\theta=\frac v r$.

The $r\ddot\theta$ term is new. It points along the path of the particle and together with the $-r\dot\theta^2$ term it makes sures the radius stays constant. Note that if $\dot\theta$=constant this term drops out and we have regular old uniform circular motion.


EDIT I'll provide a slightly easier to read explanation.

In circular motion the velocity of a particle is always at 90 degrees with its radius. Like in this picture

circular motion

When the speed says the same we have uniform circular motion. In that case the acceleration point directly at the center. We can decompose the acceleration into two components: one that points towards the center called 'centripetal acceleration' or $a_c$ and one that points along the velocity of the particle called 'tangential acceleration' or $a_t$.

In uniform circular motion we have $a_t=0$ since the acceleration is only towards the center. Also the centipetal acceleration is given by $a_c=\frac{v^2}r$.

If we accelerate in the tangential direction the speed of the particle increases. In that case the centripetal acceleration must increase to compensate, because $a_c=\frac{v^2}r$ and $r$ is constant. It's easier to see if you write it as $r=\frac{v^2}{a_c}$. If we make the speed twice as large then $a_c$ must get 4 times as big.

So long story short, it is possible to accelerate in the tangential direction but to do that you must increase the centripetal acceleration in a precise way to keep the same radius. Similarly we must decrease $a_c$ if we decelerate in the tangential direction.

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  • $\begingroup$ This getting a bit complicated for me to understand sir.... $\endgroup$
    – PATRICK
    Commented Sep 30, 2020 at 12:16
  • $\begingroup$ Ah sorry it's hard to know sometimes at what level you should explain things. I added an extra section that's a little easier to read. $\endgroup$ Commented Sep 30, 2020 at 12:51
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During non uniform circular motion,the direction of net acceleration is not in the direction of the centripetal acceleration, then why does a particle still move in a circular path,please explain..

Assuming by "non uniform circular motion" you mean a change in speed of the particle moving in a circle, then it is because the centripetal acceleration depends only on the magnitude of the tangential velocity (the speed of the particle), not on the rate of change of speed of the particle, or change in tangential velocity (tangential acceleration). The following explanation is offered:

For circular motion there are two types of possible acceleration: centripetal and tangential.

Centripetal acceleration, $a_c$, is the acceleration towards the center of the circular path. It is always present and it is what keeps the particle in circular motion. It is due to a centripetal force. In the case of a car, the centripetal force is the static friction force between the tires and the road and directed towards the center of the circular path. The centripetal acceleration depends on the magnitude of the tangential velocity $v_t$ (the speed of the car or its angular velocity, ω, in rad/s) and the radius $r$ of the circular motion according to,

$$a_{c}=\frac{v^{2}_t}{r}=rω^2$$

The tangential acceleration $a_t$ results from the change in the magnitude of the tangential velocity. An object can move in a circle and not have any tangential acceleration simply because the angular acceleration $α$ (rad/sec$^2$) is zero because the object is moving with a constant angular velocity ω ($\Delta ω =0$). In the case of a car in circular motion this is the acceleration due to braking or increasing the speed of the car due to the static friction force between the tires and the road in the tangential direction.

$$a_{t}=rα=\frac{\Delta ω}{\Delta t}$$

Hope this helps.

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