Question:

A car is moving on a circular path of radius \(600\ m\) such that the magnitudes of the tangential acceleration and centripetal acceleration are equal. The time taken by the car to complete first quarter of revolution, if it is moving with an initial speed of \(54 \ km/hr\) is \(𝑑(1βˆ’π‘’^{βˆ’\frac \pi2})\ 𝑠\). The value of \(t\) is _____ .

Updated On: Mar 20, 2025
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Correct Answer: 40

Solution and Explanation

Step 1: Relating Tangential and Centripetal Acceleration
Given: \[ v_0 = 54 \, \text{km/hr} = 15 \, \text{m/s}, \quad R = 600 \, \text{m}. \] The tangential acceleration is \(a_t = \frac{dv}{dt}\), and the centripetal acceleration is \(a_c = \frac{v^2}{R}\). The problem states that: \[ a_t = a_c \implies \frac{dv}{dt} = \frac{v^2}{R}. \] Step 2: Differential Equation for Velocity
Rearrange the equation: \[ \frac{dv}{v^2} = \frac{dt}{R}. \] Integrate both sides: \[ \int_{v_0}^{v} \frac{dv}{v^2} = \int_{0}^{t} \frac{dt}{R}. \] The integration gives: \[ -\frac{1}{v} \Big|_{v_0}^v = \frac{t}{R}. \] Simplify: \[ \frac{1}{v} - \frac{1}{v_0} = \frac{t}{R}. \] Step 3: Solve for Velocity as a Function of Time
Rearrange to find \(v(t)\): \[ \frac{1}{v} = \frac{1}{v_0} + \frac{t}{R}. \] Thus: \[ v(t) = \frac{1}{\frac{1}{v_0} + \frac{t}{R}}. \] Step 4: Angular Displacement for Quarter Revolution
The angular displacement \(\theta\) for circular motion is related to velocity by: \[ d\theta = \frac{v}{R} dt. \] Substitute \(v(t)\): \[ d\theta = \frac{1}{R \left( \frac{1}{v_0} + \frac{t}{R} \right)} dt. \] Integrate from \(\theta = 0\) to \(\pi/2\) for the quarter revolution: \[ \int_{0}^{\pi/2} d\theta = \int_{0}^{t} \frac{1}{R \left( \frac{1}{v_0} + \frac{t}{R} \right)} dt. \] Solve the integral: \[ \theta = -\ln \left( 1 - \frac{t R}{v_0} \right). \] For \(\theta = \pi/2\), substitute and solve: \[ t = 40 (1 - e^{-\pi/2}) \, \text{s}. \] Thus, the time for the first quarter revolution is \(t = 40 \, \text{s}\).
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Concepts Used:

Uniform Circular Motion

A circular motion is defined as the movement of a body that follows a circular route. The motion of a body going at a constant speed along a circular path is known as uniform circular motion. The velocity varies while the speed of the body in uniform circular motion remains constant.

Uniform Circular Motion Examples:

  • The motion of electrons around its nucleus.
  • The motion of blades of the windmills.

Uniform Circular Motion Formula:

When the radius of the circular path is R, and the magnitude of the velocity of the object is V. Then, the radial acceleration of the object is:

arad = v2/R

Similarly, this radial acceleration is always perpendicular to the velocity direction. Its SI unit is m2sβˆ’2.

The radial acceleration can be mathematically written using the period of the motion i.e. T. This period T is the volume of time taken to complete a revolution. Its unit is measurable in seconds.

When angular velocity changes in a unit of time, it is a radial acceleration.

Angular acceleration indicates the time rate of change of angular velocity and is usually denoted by Ξ± and is expressed in radians per second. Moreover, the angular acceleration is constant and does not depend on the time variable as it varies linearly with time. Angular Acceleration is also called Rotational Acceleration.

Angular acceleration is a vector quantity, meaning it has magnitude and direction. The direction of angular acceleration is perpendicular to the plane of rotation.

Formula Of Angular Acceleration

The formula of angular acceleration can be given in three different ways.

α = dωdt

Where,

Ο‰ β†’ Angular speed

t β†’ Time

Ξ± = d2ΞΈdt2

Where,

ΞΈ β†’ Angle of rotation

t β†’ Time

Average angular acceleration can be calculated by the formula below. This formula comes in handy when angular acceleration is not constant and changes with time.

Ξ±avg = Ο‰2 - Ο‰1t2 - t1

Where,

Ο‰1 β†’ Initial angular speed

Ο‰2 β†’ Final angular speed

t1 β†’ Starting time

t2 β†’ Ending time

Also Read: Angular Motion