For a particle in uniform circular motion the acceleration a at a point P(R, theta) on the circle of

Q: Why is there no tangential acceleration?

Because the particle is in *uniform* circular motion, meaning its speed $v$ is constant. Tangential acceleration only exists when the speed changes.

Q: Why are both components negative in the first quadrant?

In the first quadrant, the position vector points right and up (positive $x$ and $y$). Since centripetal acceleration points towards the origin (opposite to the position vector), it must point left and down, making both components negative.

Q: Does this formula hold true in other quadrants?

Yes! The angle $\theta$ is measured from the positive $x$-axis. The trigonometric functions $\cos\theta$ and $\sin\theta$ will automatically adjust their signs based on the quadrant, ensuring the acceleration always points towards the center.

Q: What is the unit vector for the radial direction?

The outward radial unit vector is $\hat{\mathbf{r}} = \cos\theta\hat{\mathbf{i}} + \sin\theta\hat{\mathbf{j}}$. The inward radial unit vector is simply $-\hat{\mathbf{r}}$.

Q: Can we use position vector $\mathbf{r}$ to write the acceleration?

Absolutely! Since $\mathbf{r} = R\cos\theta\hat{\mathbf{i}} + R\sin\theta\hat{\mathbf{j}}$, we can write $\mathbf{a} = -\frac{v^2}{R^2}\mathbf{r} = -\omega^2\mathbf{r}$, where $\omega = v/R$ is the angular velocity.

Question

For a particle in uniform circular motion the acceleration $\mathbf{a}$ at a point $P(R, 	heta)$ on the circle of radius $R$ is (here, $	heta$ is measured from the $x$-axis)

Options:

(A) $-\frac{v^2}{R}\cos	heta\hat{\mathbf{i}} + \frac{v^2}{R}\sin	heta\hat{\mathbf{j}}$

(B) $-\frac{v^2}{R}\sin	heta\hat{\mathbf{i}} + \frac{v^2}{R}\cos	heta\hat{\mathbf{j}}$

(C) $-\frac{v^2}{R}\cos	heta\hat{\mathbf{i}} - \frac{v^2}{R}\sin	heta\hat{\mathbf{j}}$

(D) $\frac{v^2}{R}\hat{\mathbf{i}} + \frac{v^2}{R}\hat{\mathbf{j}}$

Sigma Percentile · Accepted Answer

### The Beauty of Uniform Circular Motion

Imagine you are standing in an open field, swinging a stone tied to a string in a perfect horizontal circle. The stone moves at a constant speed, never speeding up or slowing down. This is the classic setup of **uniform circular motion**.

At first glance, you might think, "If the speed is constant, the acceleration must be zero, right?"

But here is the catch. Velocity is a vector—it has both magnitude (speed) and direction. Even though the speed is constant, the direction of the stone is changing at every single microsecond. And in the world of physics, a change in direction means a change in velocity, which absolutely requires an acceleration!

### Decoding the Acceleration

So, where does this acceleration point?

If you were to suddenly let go of the string, the stone would fly off in a straight line tangent to the circle. The string is constantly pulling the stone *inward*, preventing it from flying away. This inward pull provides the **centripetal acceleration**.

The word "centripetal" literally means "center-seeking." This acceleration always points directly towards the center of the circular path.

For a particle moving with a constant speed $v$ in a circle of radius $R$, the magnitude of this centripetal acceleration is given by a beautifully simple formula:

$$a_c = \frac{v^2}{R}$$

This tells us *how much* acceleration there is. But to fully describe it, we need to know its exact direction in a coordinate system.

### The Art of Vector Resolution

Let's place our circle on an $x$-$y$ coordinate plane, with the center of the circle right at the origin $(0,0)$.

Suppose our particle is currently at a point $P(R, 	heta)$, where $	heta$ is the angle measured counterclockwise from the positive $x$-axis.

First, let's think about the **position vector** of the particle. It points from the origin outward to point $P$. Using basic trigonometry, the unit vector pointing in this outward radial direction is:

$$\hat{\mathbf{r}} = \cos	heta \hat{\mathbf{i}} + \sin	heta \hat{\mathbf{j}}$$

Now, remember our centripetal acceleration? It points exactly in the *opposite* direction—from point $P$ inward towards the origin.

Therefore, the direction of our acceleration vector is simply the negative of the outward radial unit vector:

$$	ext{Direction} = -\hat{\mathbf{r}} = -\cos	heta \hat{\mathbf{i}} - \sin	heta \hat{\mathbf{j}}$$

### The Final Masterpiece

We have the magnitude, and we have the direction. In physics, constructing a vector is as simple as multiplying its magnitude by its direction unit vector.

Let's put it all together:

$$\mathbf{a} = |\mathbf{a}_c| (-\hat{\mathbf{r}})$$

Substitute the magnitude we found earlier:

$$\mathbf{a} = \frac{v^2}{R} (-\cos	heta \hat{\mathbf{i}} - \sin	heta \hat{\mathbf{j}})$$

Now, just distribute the magnitude into the parentheses to get the final $x$ and $y$ components:

$$\mathbf{a} = -\frac{v^2}{R}\cos	heta \hat{\mathbf{i}} - \frac{v^2}{R}\sin	heta \hat{\mathbf{j}}$$

And there we have it!

Notice how both components are negative when the particle is in the first quadrant. This makes perfect physical sense because the acceleration vector must point left (negative $x$) and down (negative $y$) to aim directly at the origin.

This elegant expression perfectly matches option (c). It is a brilliant example of how physical intuition and mathematical resolution work hand-in-hand to describe the universe!

Animated Solution for Physics - Kinematics: For a particle in uniform circular motion the acceleration $a$ at a point $P (R, θ)$ on the circle of radius $R$ is (here, $θ$ is measured from the $x$ -axis)

Select Answer:

Visualized Solution

$Visualizing the Circular Path$

$Centripetal Acceleration$

$Resolving into Components$

$Vector Formulation$

$Final Expression$

$Non-Uniform Circular Motion$

The Sigma Insight: Kinematics of Circular Motion

The Beauty of Uniform Circular Motion

Decoding the Acceleration

The Art of Vector Resolution

The Final Masterpiece

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