Animated Solution for Physics - Kinematics: Spotlight S rotates in a horizontal plane with constant angular velocity of 0.1 rad/s. The spot of light P moves along the wall at a distance of 3 m. The velocity of the spot P when θ=45∘ (see fig.) is……. m/s.
Enter Numerical Value:
Visualized Solution
Visualizing the Setup
Let the spotlight be at point S.
The light beam hits the wall at a moving spot P.
Defining the Geometry
Drop a perpendicular from S to the wall. The distance is 3 m.
Let x be the distance of spot P from the perpendicular foot.
Let θ be the angle the beam makes with the perpendicular.
Relating Position and Angle
In the right-angled triangle, use the tangent function.
tanθ=AdjacentOpposite=3x
Equation for Position x
Rearrange to express x in terms of θ.
x=3tanθ
Defining Velocities
Linear velocity of the spot: vP=dtdx
Angular velocity of the spotlight: ω=dtdθ
Differentiating the Position Equation
Differentiate x=3tanθ with respect to time t.
Applying the Chain Rule
dtdx=3⋅dtd(tanθ)
dtdx=3sec2θ⋅dtdθ
Substituting Known Variables
Substitute dtdx=vP and dtdθ=ω.
vP=3sec2θ⋅ω
Plugging in the Given Values
Given: ω=0.1 rad/s and θ=45∘.
vP=3sec2(45∘)⋅(0.1)
Evaluating sec(45∘)
Recall that sec(45∘)=2.
Therefore, sec2(45∘)=(2)2=2.
Calculating Final Velocity
Substitute the evaluated trigonometric term back into the equation.
vP=3⋅2⋅0.1
Conclusion
vP=0.6 m/s
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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 S. We drop a perpendicular line straight from the spotlight to the wall. The problem states that this shortest distance is exactly 3 m.
Now, let's look at the moving spot of light on the wall, which we will call P. The distance from the foot of our perpendicular line to the spot P is our horizontal position, x. Finally, the light beam itself forms the hypotenuse of a right-angled triangle. The angle between the perpendicular line and the light beam is θ.
We now have a perfect right-angled triangle where the adjacent side is 3 m, the opposite side is x, and the angle is θ.
The Master Equation
Our goal is to find the velocity of the spot, which means we need an equation that describes its position x. Since we have a right-angled triangle, trigonometry is our best tool. We need a ratio that connects the opposite side (x) and the adjacent side (3). The tangent function is exactly what we need.
We can write the relationship as:
tanθ=3x
To make this equation useful for kinematics, we need to isolate the position variable x. Multiplying both sides by 3, we get our master equation for the position of the spot:
x=3tanθ
The Calculus of Motion
Now we transition from static geometry to dynamic kinematics. The spot is moving, meaning its position x is changing with time t. The rate of change of position is the linear velocity, vP=dtdx. Simultaneously, the spotlight is rotating, meaning the angle θ is changing with time. The rate of change of the angle is the angular velocity, ω=dtdθ.
To find the linear velocity, we must differentiate our master equation with respect to time t. This is where many students fall into a trap. We are differentiating with respect to time, not θ! Therefore, we must employ the chain rule from calculus.
Differentiating both sides of x=3tanθ with respect to t yields:
dtdx=3⋅dtd(tanθ)
Applying the chain rule, the derivative of tanθ is sec2θ, multiplied by the derivative of the inner function θ with respect to t:
dtdx=3sec2θ⋅dtdθ
Now, we substitute our kinematic variables back into the equation. Replacing dtdx with vP and dtdθ with ω, we get a beautiful formula linking linear and angular velocity:
vP=3sec2θ⋅ω
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 θ=45∘. We are also given that the constant angular velocity ω=0.1 rad/s.
Substituting these values into our derived formula:
vP=3sec2(45∘)⋅(0.1)
We know from basic trigonometry that cos(45∘)=21. Therefore, its reciprocal, sec(45∘), is 2. Squaring this value gives us sec2(45∘)=2.
Substituting this back into our calculation:
vP=3⋅2⋅0.1
vP=6⋅0.1=0.6 m/s
The velocity of the spot of light on the wall at that specific instant is 0.6 m/s. 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.