A particle move In plane a/c to X=Rsinwt+Rwt and y=Rcoswt wher wand r are constant .find instantaneous acceleration and velocity at min and max value of y

To analyze the motion of the particle described by the equations \( x = R \sin(\omega t) + R \omega t \) and \( y = R \cos(\omega t) \), we need to find the instantaneous velocity and acceleration at the maximum and minimum values of \( y \). Let's break this down step by step.

Understanding the Motion

The equations represent a particle moving in a plane, where \( R \) is a constant amplitude, \( \omega \) is the angular frequency, and \( t \) is time. The \( y \) component, given by \( y = R \cos(\omega t) \), oscillates between \( -R \) and \( R \). The maximum value of \( y \) occurs when \( \cos(\omega t) = 1 \) (i.e., \( y = R \)), and the minimum value occurs when \( \cos(\omega t) = -1 \) (i.e., \( y = -R \)).

Finding the Maximum and Minimum Values of y

• Maximum \( y \): \( y = R \) when \( \omega t = 0, 2\pi, 4\pi, \ldots \)
• Minimum \( y \): \( y = -R \) when \( \omega t = \pi, 3\pi, 5\pi, \ldots \)

Calculating Instantaneous Velocity

The instantaneous velocity \( \mathbf{v} \) of the particle can be found by differentiating the position equations with respect to time \( t \).

Velocity Components

We differentiate \( x \) and \( y \) as follows:

• For \( x \):

  \( v_x = \frac{dx}{dt} = R \omega \cos(\omega t) + R \omega \)

• For \( y \):

  \( v_y = \frac{dy}{dt} = -R \omega \sin(\omega t) \)

Velocity at Maximum y

At maximum \( y = R \) (when \( \omega t = 0 \)):

• \( v_x = R \omega \cos(0) + R \omega = 2R \omega \)
• \( v_y = -R \omega \sin(0) = 0 \)

Thus, the velocity at maximum \( y \) is \( \mathbf{v} = (2R \omega, 0) \).

Velocity at Minimum y

At minimum \( y = -R \) (when \( \omega t = \pi \)):

• \( v_x = R \omega \cos(\pi) + R \omega = -R \omega + R \omega = 0 \)
• \( v_y = -R \omega \sin(\pi) = 0 \)

Thus, the velocity at minimum \( y \) is \( \mathbf{v} = (0, 0) \).

Calculating Instantaneous Acceleration

The instantaneous acceleration \( \mathbf{a} \) is found by differentiating the velocity components with respect to time.

Acceleration Components

• For \( x \):

  \( a_x = \frac{d^2x}{dt^2} = -R \omega^2 \sin(\omega t) \)

• For \( y \):

  \( a_y = \frac{d^2y}{dt^2} = -R \omega^2 \cos(\omega t) \)

Acceleration at Maximum y

At maximum \( y = R \) (when \( \omega t = 0 \)):

• \( a_x = -R \omega^2 \sin(0) = 0 \)
• \( a_y = -R \omega^2 \cos(0) = -R \omega^2 \)

Thus, the acceleration at maximum \( y \) is \( \mathbf{a} = (0, -R \omega^2) \).

Acceleration at Minimum y

At minimum \( y = -R \) (when \( \omega t = \pi \)):

• \( a_x = -R \omega^2 \sin(\pi) = 0 \)
• \( a_y = -R \omega^2 \cos(\pi) = R \omega^2 \)

Thus, the acceleration at minimum \( y \) is \( \mathbf{a} = (0, R \omega^2) \).

Summary of Results

To summarize:

• At maximum \( y \) (\( y = R \)):
  • Velocity: \( \mathbf{v} = (2R \omega, 0) \)
  • Acceleration: \( \mathbf{a} = (0, -R \omega^2) \)
• At minimum \( y \) (\( y = -R \)):
  • Velocity: \( \mathbf{v} = (0, 0) \)
  • Acceleration: \( \mathbf{a} = (0, R \omega^2) \)

This analysis shows how the particle behaves at the extremes of its vertical motion, providing insights into both its velocity and acceleration at those points. If you have any further questions or need clarification on any part, feel free to ask!

To analyze the motion of the particle described by the equations \( x = R \sin(\omega t) + R \omega t \) and \( y = R \cos(\omega t) \), we need to find the instantaneous velocity and acceleration at the maximum and minimum values of \( y \). Let's break this down step by step.

Understanding the Motion

The equations given represent a particle moving in a plane. The \( y \)-coordinate is defined by \( y = R \cos(\omega t) \), which indicates that \( y \) oscillates between \( -R \) and \( R \). The maximum value of \( y \) occurs when \( \cos(\omega t) = 1 \) (i.e., \( y = R \)), and the minimum value occurs when \( \cos(\omega t) = -1 \) (i.e., \( y = -R \)).

Finding Critical Points for \( y \)

To find the times when \( y \) reaches its maximum and minimum values, we can set \( \cos(\omega t) \) to 1 and -1:

• Maximum \( y \): \( \omega t = 0, 2\pi, 4\pi, \ldots \) (i.e., \( t = \frac{2n\pi}{\omega} \) for integers \( n \))
• Minimum \( y \): \( \omega t = \pi, 3\pi, 5\pi, \ldots \) (i.e., \( t = \frac{(2n+1)\pi}{\omega} \) for integers \( n \))

Calculating Instantaneous Velocity

The instantaneous velocity \( \mathbf{v} \) is given by the derivative of the position vector with respect to time:

\( \mathbf{v} = \left( \frac{dx}{dt}, \frac{dy}{dt} \right) \)

Deriving \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \)

First, we differentiate \( x \) and \( y \):

• For \( x \):

  \( \frac{dx}{dt} = R \omega \cos(\omega t) + R \omega \)

• For \( y \):

  \( \frac{dy}{dt} = -R \omega \sin(\omega t) \)

Velocity at Maximum and Minimum \( y \)

Now, we evaluate the velocity at the critical points:

• At maximum \( y \) (\( t = \frac{2n\pi}{\omega} \)):

  \( \frac{dy}{dt} = 0 \) (since \( \sin(0) = 0 \))

  \( \frac{dx}{dt} = R \omega \) (since \( \cos(0) = 1 \))

  Thus, \( \mathbf{v}_{\text{max}} = (R \omega, 0) \)

• At minimum \( y \) (\( t = \frac{(2n+1)\pi}{\omega} \)):

  \( \frac{dy}{dt} = 0 \) (since \( \sin(\pi) = 0 \))

  \( \frac{dx}{dt} = 0 \) (since \( \cos(\pi) = -1 \) and \( R \omega - R \omega = 0 \))

  Thus, \( \mathbf{v}_{\text{min}} = (0, 0) \)

Finding Instantaneous Acceleration

The instantaneous acceleration \( \mathbf{a} \) is the derivative of the velocity:

\( \mathbf{a} = \left( \frac{d^2x}{dt^2}, \frac{d^2y}{dt^2} \right) \)

Calculating \( \frac{d^2x}{dt^2} \) and \( \frac{d^2y}{dt^2} \)

We differentiate the velocity components:

• For \( x \):

  \( \frac{d^2x}{dt^2} = -R \omega^2 \sin(\omega t) \)

• For \( y \):

  \( \frac{d^2y}{dt^2} = -R \omega^2 \cos(\omega t) \)

Acceleration at Maximum and Minimum \( y \)

Now, we evaluate the acceleration at the critical points:

• At maximum \( y \):

  \( \frac{d^2y}{dt^2} = -R \omega^2 \) (since \( \cos(0) = 1 \))

  \( \frac{d^2x}{dt^2} = 0 \) (since \( \sin(0) = 0 \))

  Thus, \( \mathbf{a}_{\text{max}} = (0, -R \omega^2) \)

• At minimum \( y \):

  \( \frac{d^2y}{dt^2} = R \omega^2 \) (since \( \cos(\pi) = -1 \))

  \( \frac{d^2x}{dt^2} = 0 \) (since \( \sin(\pi) = 0 \))

  Thus, \( \mathbf{a}_{\text{min}} = (0, R \omega^2) \)

Summary of Results

In summary, at the maximum value of \( y \) (when \( y = R \)), the instantaneous velocity is \( (R \omega, 0) \) and the acceleration is \( (0, -R \omega^2) \). At the minimum value of \( y \) (when \( y = -R \)), the instantaneous velocity is \( (0, 0) \)