Sigma Percentile
JEE Advanced 1987
LEVELJEE Main

Animated Solution for Physics - Kinematics: Spotlight rotates in a horizontal plane with constant angular velocity of . The spot of light moves along the wall at a distance of . The velocity of the spot when (see fig.) is……. .

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

  • Let the spotlight be at point .
  • The light beam hits the wall at a moving spot .

Defining the Geometry

  • Drop a perpendicular from to the wall. The distance is .
  • Let be the distance of spot from the perpendicular foot.
  • Let be the angle the beam makes with the perpendicular.

Relating Position and Angle

  • In the right-angled triangle, use the tangent function.

Equation for Position

  • Rearrange to express in terms of .

Defining Velocities

  • Linear velocity of the spot:
  • Angular velocity of the spotlight:

Differentiating the Position Equation

  • Differentiate with respect to time .

Applying the Chain Rule

Substituting Known Variables

  • Substitute and .

Plugging in the Given Values

  • Given: and .

Evaluating

  • Recall that .
  • Therefore, .

Calculating Final Velocity

  • Substitute the evaluated trigonometric term back into the equation.

Conclusion

The Sigma Insight: Motion in a Plane

Solution Diagram
The Dance of Light: A Journey into Related Rates
Imagine standing in a dark room, holding a laser pointer. You shine it directly onto a straight wall in front of you. Now, you start rotating your wrist at a steady, constant pace. What happens to the bright spot of light on the wall? It races across the surface. But here is the fascinating part: even though your wrist is turning at a constant speed, the spot on the wall does not move at a constant speed. It actually accelerates! This beautiful interplay between rotational motion and linear motion is a classic physics concept known as 'related rates'. Let's dive into the mathematics behind this phenomenon.

Analyzing the Setup

To understand the motion, we first need to freeze time and build a geometric model of our physical reality. Let's define the position of our spotlight as point . We drop a perpendicular line straight from the spotlight to the wall. The problem states that this shortest distance is exactly .
Now, let's look at the moving spot of light on the wall, which we will call . The distance from the foot of our perpendicular line to the spot is our horizontal position, . Finally, the light beam itself forms the hypotenuse of a right-angled triangle. The angle between the perpendicular line and the light beam is .
We now have a perfect right-angled triangle where the adjacent side is , the opposite side is , and the angle is .

The Master Equation

Our goal is to find the velocity of the spot, which means we need an equation that describes its position . Since we have a right-angled triangle, trigonometry is our best tool. We need a ratio that connects the opposite side () and the adjacent side (). The tangent function is exactly what we need.
We can write the relationship as:
To make this equation useful for kinematics, we need to isolate the position variable . Multiplying both sides by , we get our master equation for the position of the spot:

The Calculus of Motion

Now we transition from static geometry to dynamic kinematics. The spot is moving, meaning its position is changing with time . The rate of change of position is the linear velocity, . Simultaneously, the spotlight is rotating, meaning the angle is changing with time. The rate of change of the angle is the angular velocity, .
To find the linear velocity, we must differentiate our master equation with respect to time . This is where many students fall into a trap. We are differentiating with respect to time, not ! Therefore, we must employ the chain rule from calculus.
Differentiating both sides of with respect to yields:
Applying the chain rule, the derivative of is , multiplied by the derivative of the inner function with respect to :
Now, we substitute our kinematic variables back into the equation. Replacing with and with , we get a beautiful formula linking linear and angular velocity:

Final Calculation

With our formula ready, we can plug in the specific values given in the problem. We need to find the velocity at the exact instant when the angle . We are also given that the constant angular velocity .
Substituting these values into our derived formula:
We know from basic trigonometry that . Therefore, its reciprocal, , is . Squaring this value gives us .
Substituting this back into our calculation:
The velocity of the spot of light on the wall at that specific instant is . This problem beautifully demonstrates how calculus acts as a bridge between static geometry and dynamic motion, allowing us to predict the exact behavior of moving systems.

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