A straight track is tangent to a circular track of radius r. Two material points A and B start simul

Q: Why does point B have tangential acceleration if point A moves with uniform velocity?

Even though A's linear velocity is constant, the angular velocity of the line connecting them is not constant. Since B's speed depends on this changing angular velocity ($v_B = r\omega$), B's speed is changing, which means it must have a tangential acceleration.

Q: Can I just use $a = r\alpha$ for the total acceleration?

No, $r\alpha$ only gives the tangential component of acceleration. Because B is moving in a circle, it also experiences a centripetal (normal) acceleration directed towards the center. You must account for both.

Q: How did we get the velocity of B?

Point B is constrained to move on a circle of radius $r$. Its linear speed is always given by the product of its radial distance and its angular velocity, $v = r\omega$. Since B is collinear with A, they share the same $\omega$.

Q: Why is the tangential acceleration negative?

The negative sign indicates that as the angle $\theta$ increases, the angular velocity $\omega$ decreases. Consequently, the linear speed of point B is decreasing over time, meaning it is decelerating along its path.

Q: What happens to the acceleration when $\theta$ is zero?

When $\theta = 0$, point A is exactly at the point of tangency. The formula gives $a = u^2/r$. At this instant, the tangential acceleration is zero (since $\sin(0)=0$), and the entire acceleration is purely centripetal.

Question

A straight track is tangent to a circular track of radius $r$. Two material points A and B start simultaneously from the common point of the tracks. The point A moves with uniform velocity $u$ on the straight track whereas the point B on the circular track always keeping itself collinear with the centre of the circular track and the point A. Find suitable expression for magnitude of acceleration of the point B when it is at angular position $	heta$.

Sigma Percentile · Accepted Answer

### Setting the Scene: A Tale of Two Tracks

Imagine a perfectly straight highway running tangent to a massive circular test track. Two vehicles, A and B, start at the exact point where the straight track touches the circle. Vehicle A speeds off down the straight highway at a constant velocity $u$. Vehicle B, however, is bound to the circular track. But there's a catch: a powerful invisible laser beam connects the center of the circle, vehicle B, and vehicle A at all times. This means they are strictly collinear.

Our mission is to find the total acceleration of vehicle B when the laser beam makes an angle $	heta$ with the vertical.

### The Kinematic Constraint: Finding the Angular Velocity

Let's establish the geometry. If the circle has a radius $r$, and the angle is $	heta$, the distance vehicle A has traveled along the straight track is simply the opposite side of a right-angled triangle.

We can write this as:
$$x = r 	an	heta$$

Since vehicle A moves with a constant velocity $u$, we know that $u = \frac{dx}{dt}$. Let's differentiate our geometric constraint with respect to time to see how fast the angle is changing:
$$u = \frac{d}{dt}(r 	an	heta) = r \sec^2	heta \cdot \frac{d	heta}{dt}$$

The term $\frac{d	heta}{dt}$ is the angular velocity $\omega$ of the invisible laser beam. Rearranging for $\omega$, we get:
$$\omega = \frac{u}{r \sec^2	heta} = \frac{u}{r} \cos^2	heta$$

This is a crucial realization: even though A moves at a constant speed, the angular velocity $\omega$ is **not constant**. It decreases as $	heta$ increases.

### The Motion of Point B

Because vehicle B is trapped on the laser beam, it shares this exact same angular velocity $\omega$. However, B is also constrained to move on a circle of constant radius $r$.

The linear speed of B is therefore:
$$v_B = r\omega = r \left(\frac{u}{r} \cos^2	hetaight) = u \cos^2	heta$$

### The Dual Nature of Acceleration

Because B is moving on a curved path with a changing speed, it experiences two distinct types of acceleration:
1.  **Tangential Acceleration ($a_t$)**: Due to the changing magnitude of its velocity.
2.  **Normal/Centripetal Acceleration ($a_n$)**: Due to the changing direction of its velocity.

Let's calculate the tangential component by differentiating $v_B$ with respect to time. We must use the chain rule:
$$a_t = \frac{dv_B}{dt} = \frac{d}{dt}(u \cos^2	heta) = u(2\cos	heta)(-\sin	heta)\frac{d	heta}{dt}$$

Substituting $\omega$ for $\frac{d	heta}{dt}$:
$$a_t = -2u \sin	heta \cos	heta \left(\frac{u}{r} \cos^2	hetaight) = -\frac{2u^2}{r} \sin	heta \cos^3	heta$$

The negative sign perfectly aligns with our intuition: as the angle grows, B slows down.

Next, we calculate the normal acceleration, which always points towards the center of the circle:
$$a_n = \frac{v_B^2}{r} = r\omega^2 = r \left(\frac{u}{r} \cos^2	hetaight)^2 = \frac{u^2}{r} \cos^4	heta$$

### The Grand Finale: Total Acceleration

Since the tangential and normal accelerations are perpendicular vectors, we find the magnitude of the total acceleration using the Pythagorean theorem:
$$a = \sqrt{a_t^2 + a_n^2}$$

Substituting our derived components:
$$a = \sqrt{\left(-\frac{2u^2}{r} \sin	heta \cos^3	hetaight)^2 + \left(\frac{u^2}{r} \cos^4	hetaight)^2}$$

Let's factor out the common terms to simplify this beast:
$$a = \frac{u^2}{r} \cos^3	heta \sqrt{4 \sin^2	heta + \cos^2	heta}$$

Using the trigonometric identity $\cos^2	heta = 1 - \sin^2	heta$, we can refine the expression under the square root:
$$4 \sin^2	heta + 1 - \sin^2	heta = 1 + 3 \sin^2	heta$$

This leads us to our final, elegant result:
**$$a = \frac{u^2}{r} \cos^3	heta \sqrt{1 + 3 \sin^2	heta}$$**

Visualized Solution

$x = r tan θ$

$u = \frac{d x}{d t}$

$ω = \frac{u}{r} cos^{2} θ$

$v_{B} = u cos^{2} θ$

$a_{t} = \frac{d v _{B}}{d t}$

$a_{t} = - \frac{2 u ^{2}}{r} sin θ cos^{3} θ$

$a_{n} = \frac{u ^{2}}{r} cos^{4} θ$

$a = a_{t}^{2} + a_{n}^{2}$

$a = \frac{u ^{2}}{r} cos^{3} θ 1 + 3 sin^{2} θ$

The Sigma Insight: Kinematics of Circular Motion

Setting the Scene

The Kinematic Constraint

The Motion of Point B

The Dual Nature of Acceleration

The Grand Finale

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=rtanθ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } u=dtdx​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ω=ru​cos2θ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } vB​=ucos2θ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } at​=dtdvB​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } at​=−r2u2​sinθcos3θ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } an​=ru2​cos4θ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a=at2​+an2​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a=ru2​cos3θ1+3sin2θ​