Centripetal Force For Non-uniform Circular Motion

In non-uniform circular motion, where an object moves along a circular path with varying speed, the concept of centripetal force becomes crucial. Unlike uniform circular motion, where speed remains constant, non-uniform circular motion involves acceleration or deceleration, requiring a varying centripetal force to maintain the curved trajectory. This force continuously acts towards the center of the circle, adjusting to the changing velocity of the object. A real-life example can be seen in a car navigating a curved road. As the driver accelerates or brakes, the force keeping the car on the road's curve varies, ensuring it doesn't skid off the path. Understanding centripetal force in non-uniform circular motion is essential for predicting and managing such dynamic situations.

Centripetal Force for Non-Uniform Circular Motion

$\begin{aligned} & F_c=m a_c=\frac{m v^2}{r} \quad\left(\vec{F}_c \perp \vec{v}\right) \\ & \mathrm{F}_{\mathrm{t}}=m \mathrm{a}_{\mathrm{t}} \\ & F_{n e t}=m \sqrt{a_c^2+a_t^2} \\ & \mathrm{~m}=\text { mass } \\ & \mathrm{a}_{\mathrm{c}}=\text { centripetal acceleration } \\ & \mathrm{a}_{\mathrm{t}}=\text { tangential acceleration } \\ & \mathrm{F}_{\mathrm{c}}=\text { centripetal force }\end{aligned}$

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Solved Examples Based on Centripetal Force For Non-uniform Circular Motion

Example 1: Which relation is always correct

where

$a_t=$ tangential acceleration
$\mathrm{F}_{\mathrm{c}}=$ centripetal force
$\mathrm{F}_{\text {net }}=$ Net force
1) $\vec{F}_c \perp \vec{V}_t$
2) $\vec{F}_t \perp \vec{V}_t$
3) $\vec{F}_{n e t} \perp \vec{V}_t$
4) All of these

Solution:

Force in non-uniform Circular Motion

$\begin{aligned} & F_c=m a_c=\frac{m v^2}{r} \\ & \mathrm{~F}_{\mathrm{t}}=m \mathrm{a}_{\mathrm{t}} \\ & F_{n e t}=m \sqrt{a_c^2+a_t^2} \\ & \mathrm{~m}=\text { mass } \\ & \mathrm{a}_{\mathrm{c}}=\text { centripetal acceleration } \\ & \mathrm{a}_{\mathrm{t}}=\text { tangential acceleration } \\ & \mathrm{F}_{\mathrm{c}}=\text { centripetal force }\end{aligned}$

As we know acceleration is always perpendicular to the direction of velocity, so, we can say that centripetal acceleration is perpendicular to the circle.

Now, from Newton's second law of motion i.e. $F_c=m a_c$, we can say that the direction of force is in the direction of acceleration or parallel to it.
So, here also the direction of the centripetal force will be parallel to centripetal acceleration. Hence, it can be said that the direction of centripetal force is perpendicular to the circle of curvature of the path and acts towards the centre as its direction of centripetal acceleration acts perpendicular to the tangential velocity, $\vec{F}_c \perp \vec{V}_t$.

Hence, the answer is the option (1).

Example 2: A car is moving with a speed of 30m/s on a circular path of 300m. Its speed is increasing at the rate of 4m/s². What is the acceleration (in m/s²) of the car?

1) 5

2) 4

3) 3

4) 10

Solution:

Force in non-uniform Circular Motion
$
\begin{aligned}
& F_c=m a_c=\frac{m v^2}{r} \quad\left(\vec{F}_c \perp \vec{v}\right) \\
& \mathrm{F}_{\mathrm{t}}=m \mathrm{a}_{\mathrm{t}} \\
& F_{\text {net }}=m \sqrt{a_c^2+a_t^2} \\
& \mathrm{~m}=\text { mass } \\
& \mathrm{a}_{\mathrm{c}}=\text { centripetal acceleration } \\
& \mathrm{a}_{\mathrm{t}}=\text { tangential acceleration } \\
& \mathrm{F}_{\mathrm{c}}=\text { centripetal force }
\end{aligned}
$
So

$
\begin{aligned}
& a_c=\frac{v^2}{R}=\frac{30 * 30}{300}=3 \mathrm{~m} / \mathrm{s}^2 \\
& a_t=4 \mathrm{~m} / \mathrm{s}^2 \\
& a=\sqrt{a_c^2+a_t^2}=\sqrt{3^2+4^2}=5 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
$

Hence, the answer is option (1).

Example 3: A car is moving with a constant speed of $20 \mathrm{~m} / \mathrm{s}_{\text {in a }}$ a circular horizontal track of radius 40 m . A bob is suspended from the roof of the car by a massless string. The angle made by the string with the vertical will be: (Take $\mathrm{g}=$ $10 \mathrm{~m} / \mathrm{s}^2$ )

1) $\frac{\pi}{3}$
2) $\frac{\pi}{2}$
3) $\frac{\pi}{4}$
4) $\frac{\pi}{6}$

Solution:

$
\begin{aligned}
& \mathrm{T} \cos \theta=\mathrm{mg}^{-(\mathrm{i})} \\
& T \sin \theta=\frac{m v^2}{R} \\
& \frac{e q(i)}{e q(i i)} ; \frac{\cos \theta}{\sin \theta}=\frac{g R}{\mathrm{v}^2} \\
& \tan \theta=\frac{\mathrm{v}^2}{\mathrm{Rg}}=\frac{400}{40 \times 10}=1 \\
& \theta=\frac{\pi}{4}
\end{aligned}
$

Hence, the answer is option (3).

Example 4: A stone of mass 900 g is tied to a string and moved in a vertical circle of radius 1 m making 10rpm. The tension in the string, when the stone is at the lowest point is (if π2=9.8 and g=9.8 m/s2 ):

1) 17.8 N

2) 8.82 N

3) 97 N

4) 9.8 N

Solution:

At a lowermost point,

$
\begin{aligned}
& T-m g=m \omega^2 r \\
& \Rightarrow T=m g+m \omega^2 r=m g+m \times 4 \pi^2 f^2 r=m g\left(1+4 f^2 r\right) \\
& \Rightarrow T=m g\left(1+4 \times \frac{100}{3600} \times 1\right)=0.9 \times 9.8 \times\left(1+\frac{1}{9}\right)=9.8 \mathrm{~N}
\end{aligned}
$

Hence, the answer is option (4).

Summary

In a non-uniform circular motion, the centripetal force varies to maintain the object's curved path as its speed changes. This force is directed towards the centre of the circle and is crucial for handling situations like a car navigating a curve. The relationship between centripetal force, tangential acceleration, and net force is key to solving problems related to objects in circular motion, as demonstrated in the examples provided.