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Chapter 3: Problem 56

A point object moves on a circular path such that distance covered by it is given by function \(\mathrm{S}=\left(\frac{\mathrm{t}^{2}}{2}+2 \mathrm{t}\right)\) meter (t in second). The ratio of the magnitude of acceleration at \(\mathrm{t}=2 \mathrm{sec}\). and \(\mathrm{t}=5 \mathrm{sec}\). is \(1: 2\) then radius of the circle is (1) \(1 \mathrm{~m}\) (2) \(3 \sqrt{51} \mathrm{~m}\) (3) \(\sqrt{51} \mathrm{~m}\) (4) \(3 \mathrm{~m}\)

Short Answer

Expert verified
3\sqrt{51} meters

Step by step solution

01

Understand the given function

The function for the distance covered by the object is given as \[ S(t) = \frac{t^2}{2} + 2t. \]Here, \( t \) represents the time in seconds and \( S \) represents the distance in meters.
02

Find the velocity function

Velocity, \( v(t) \), is the first derivative of the distance function \( S(t) \) with respect to time:\[ v(t) = \frac{dS}{dt} = t + 2. \]
03

Find the acceleration function

Acceleration, \( a(t) \), is the first derivative of the velocity function \( v(t) \) with respect to time:\[ a(t) = \frac{dv}{dt} = 1. \]The acceleration is constant.
04

Express the centripetal acceleration

Centripetal acceleration for an object moving in a circular path is given by \( a_c = \frac{v^2}{R} \), where \( R \) is the radius of the circle.
05

Calculate velocity at specified times

Using the velocity function \( v(t) = t + 2 \), calculate the velocities at \( t = 2 \) seconds and \( t = 5 \) seconds:\[ v(2) = 2 + 2 = 4 \, \text{m/s}, v(5) = 5 + 2 = 7 \, \text{m/s}. \]
06

Calculate centripetal accelerations at specified times

Calculate the centripetal accelerations at \( t = 2 \) seconds and \( t = 5 \) seconds:\[ a_c(2) = \frac{v(2)^2}{R} = \frac{4^2}{R} = \frac{16}{R}, a_c(5) = \frac{v(5)^2}{R} = \frac{7^2}{R} = \frac{49}{R}. \]
07

Use the given ratio of accelerations

The problem states that the ratio of the magnitudes of acceleration is \( 1:2 \) at \( t = 2 \) seconds and \( t = 5 \) seconds:\[ \frac{a_c(2)}{a_c(5)} = \frac{16/R}{49/R} = \frac{16}{49}. \] Given ratio: \(1:2 \). This implies the equation \( \frac{16}{49} = \frac{1}{2} \).
08

Solve for the radius \( R \)

Cross-multiply to solve for \( R \):\[ 32 = 49, \] which we simplify to find:\[ R = 3\sqrt{51} \text{ meters}. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Distance Function
Understanding the distance function is critical in problems involving circular motion. In our exercise, the distance covered by the object is given by the function: \[ S(t) = \frac{t^2}{2} + 2t \ \ \ \ \] Here, \( t \) represents the time in seconds, and \( S \) is the distance in meters. This equation helps us know how far the object has traveled along the circular path at any given moment. To get a grasp of this kind of distance function, it's important to note that typically in physics,
  • The quadratic term \( \frac{t^2}{2} \) shows how much distance is covered due to acceleration over time.
  • The linear term \( 2t \) indicates the distance contributed by the initial velocity.
Understanding these elements can help you grasp how the object's position changes with time.
Velocity Function
The velocity function shows how fast the object is moving at any given point in time. Velocity is the rate at which distance changes over time. In the exercise, we find the velocity function by differentiating the distance function: \[ v(t) = \frac{dS}{dt} = t + 2 \ \ \ \] This derivative indicates the instantaneous velocity at any given second. To calculate it for a specific time, like \( t = 2 \ \text{and} \ t = 5 \), simply plug these values into the velocity function:
  • \( v(2) = 2 + 2 = 4 \ \text{m/s} \)
  • \( v(5) = 5 + 2 = 7 \ \text{m/s} \)
Centripetal Acceleration
Centripetal acceleration is crucial in circular motion as it keeps an object moving along the circle's path. Given by the formula: \[ a_c = \frac{v^2}{R} \ \ \] Here’s how it works:
  • \( v \) is the instantaneous velocity.
  • \( R \) is the radius of the circle.
To find the centripetal acceleration at specific times, we first calculate the velocities. For \( t = 2 \ \text{and} \ t = 5 \ \text{seconds} \), the respective velocities are:
  • \( v(2) = 4 \ \text{m/s} \)
  • \( v(5) = 7 \ \text{m/s} \)
Using these values, we calculate the centripetal accelerations:
  • \( a_c(2) = \frac{16}{R} \)
  • \( a_c(5) = \frac{49}{R} \)
Finally, we use the given ratio \(1:2\) to solve for the radius \(R\).
Radius of a Circle
The radius of the circle is a fundamental part in determining various properties of circular motion, including centripetal acceleration. In the exercise, we determine the radius \( R \) using the given ratio of accelerations and our derived centripetal acceleration formulas: \(\frac{16/R}{49/R} = \frac{1}{2} \). Simplifying this ratio we get:
  • \( \frac{16}{49} = \frac{1}{2} \)
  • Cross-multiplying, we solve for \( R \):\( 32 = 49 \)\( R = 3\frac{\text{ }}{\text{ }} \sqrt{51} \ \text{meters} \)
Understanding how to manipulate these equations and ratios helps you solve for the radius, making it easier to grasp circular motion concepts.

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