Problem 6 - University Physics with Modern Physics

Question

At \(t=0\) the current to a de electric motor is reversed, resulting in an angular displacement of the motor shaft given by \(\theta(t)=(250 \mathrm{rad} / \mathrm{s}) t-\left(20.0 \mathrm{rad} / \mathrm{s}^{2}\right) t^{2}-\left(1.50 \mathrm{rad} / \mathrm{s}^{3}\right) t^{3} .(\mathrm{a}) \mathrm{At}\) what time is the angular velocity of the motor shaft zero? (b) Calculate the angular acceleration at the instant that the motor shaft has zero angular velocity. (c) How many revolutions does the motor shaft turn through between the time when the current is reversed and the instant when the angular velocity is zero? (d) How fast was the motor shaft rotating at \(t=0,\) when the current was reversed? ( c) Calculate the average angular velocity for the time period from \(t=0\) to the time calculated in part (a).

Step-by-Step Solution

Answer

(a) Solve \( 250 - 40t - 4.5t^2 = 0 \) for \( t \), (b) Substitute \( t \) into \( \alpha(t) = -40 - 9t \), (c) Find \( \theta(t) \) and convert to revolutions, (d) \( 250 \text{ rad/s} \), (e) Calculate \( \overline{\omega} \).

Step 1: Determine angular velocity expression

Angular velocity \( \omega(t) \) is the first derivative of angular displacement \( \theta(t) \). The given \( \theta(t) = 250t - 20t^2 - 1.5t^3 \). Thus, \( \omega(t) = \frac{d\theta}{dt} = 250 - 40t - 4.5t^2 \).

Step 2: Find time when angular velocity is zero

Set \( \omega(t) = 0 \) to find the time when the angular velocity is zero: \[ 250 - 40t - 4.5t^2 = 0 \] This is a quadratic equation in the form \( at^2 + bt + c = 0 \), where \( a = -4.5 \), \( b = -40 \), and \( c = 250 \). Using the quadratic formula:\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Plug in the values to calculate \( t \).

Step 3: Quadratic Formula Application

Substitute the values from Step 2 into the quadratic formula:\[ t = \frac{-(-40) \pm \sqrt{(-40)^2 - 4(-4.5)(250)}}{2(-4.5)} \] \[ t = \frac{40 \pm \sqrt{1600 + 4500}}{-9} \] \[ t = \frac{40 \pm \sqrt{6100}}{-9} \] Calculate \( t \) using a calculator.

Step 4: Calculate angular acceleration at zero angular velocity

Angular acceleration \( \alpha(t) \) is the first derivative of angular velocity \( \omega(t) \). Thus, \( \alpha(t) = \frac{d\omega}{dt} = -40 - 9t \). Substitute the value of \( t \) found in Step 3 to find \( \alpha(t) \) when angular velocity is zero.

Step 5: Calculate angular displacement for part (c)

Substitute the value of \( t \) from Step 3 into the original equation for \( \theta(t) \) to find the angular displacement at that time. Remember to convert radians to revolutions: 1 revolution = \( 2\pi \) radians.

Step 6: Determine initial angular velocity

Evaluate \( \omega(0) \) using the velocity expression from Step 1: \( \omega(0) = 250 - 40\times0 - 4.5\times0^2 = 250 \text{ rad/s} \).

Step 7: Calculate average angular velocity

The average angular velocity \( \overline{\omega} \) over the time period from \( t = 0 \) to the time found in Step 3 is:\[ \overline{\omega} = \frac{\theta(t)}{t} \]Use the angular displacement \( \theta(t) \) calculated in Step 5 and the value of \( t \) found in Step 3.

Key Concepts

Angular Displacement

Angular displacement refers to the change in the angle as an object moves along a circular path. It's expressed in radians, a unit that's vital in understanding angular motion. For an equation given like \( \theta(t) = 250t - 20t^2 - 1.5t^3 \), each term represents different parts of the motion. The coefficient of each term in the equation indicates a fixed rate at which the displacement changes over time specific to that term's power.

• The first term \( 250t \) indicates a constant rate of angular displacement.
• The second term \(-20t^2\) shows the effect of angular acceleration as it factors in the square of time.
• The third term \(-1.5t^3\) introduces a further level of complexity, indicating a change in the acceleration itself over time, common in more complex systems.

Understanding each of these helps in analyzing how the angular position evolves with time.

Angular Velocity

Angular velocity is the rate at which an object changes its angular displacement and is usually denoted as \( \omega(t) \). It's essentially the derivative of angular displacement with respect to time. This means you differentiate the equation of angular displacement to find the expression for angular velocity. For this problem, the angular velocity is:\[ \omega(t) = \frac{d\theta}{dt} = 250 - 40t - 4.5t^2 \]To find when the angular velocity is zero, set this expression to zero and solve:\[ 250 - 40t - 4.5t^2 = 0 \]Solving this quadratic equation will allow us to find the point in time when the angular speed ceases, giving a snapshot of a turning point in the system's behavior. This insight can support understanding of when motion changes direction or temporarily stops.

Angular Acceleration

Angular acceleration signifies how quickly an object's angular velocity changes over time. It is the derivative of angular velocity with respect to time, often represented by \( \alpha(t) \). For this exercise:\[ \alpha(t) = \frac{d\omega}{dt} = -40 - 9t \]To find the angular acceleration at the moment when angular velocity is zero, use the time value obtained from your calculations in the angular velocity section. Substitute the value into the equation for \( \alpha(t) \). Angular acceleration tells us how fast the angular speed is changing, providing critical information on the dynamics of the system, whether speeding up or slowing down in its rotational motion.

Quadratic Equation

A quadratic equation typically has the form \( ax^2 + bx + c = 0 \). In physics problems, such as this, they arise when calculating critical points like where velocities are zero. The quadratic equation involved here is:\[ 4.5t^2 + 40t - 250 = 0 \]To solve it, use the quadratic formula: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Plugging the values from the equation into the formula gives:

• \( a = 4.5 \)
• \( b = 40 \)
• \( c = -250 \)

These calculations provide the time instances where particular events such as angular velocity being zero occur, important for predicting motion behavior and planning in mechanical scenarios.

Revolutions

Revolutions describe the number of complete turns an object makes around a circle, directly connected to angular displacement. To understand how many revolutions occur, you need to know the total angular displacement. Since one full circle is \( 2\pi \) radians, the formula to find the number of revolutions is:\[ \,\text{Revolutions} = \frac{\theta}{2\pi} \]In our problem, first calculate the angular displacement using the equation \( \theta(t) = 250t - 20t^2 - 1.5t^3 \) with the time you've determined from the velocity equation solutions. Substitute this into the revolutions formula to know how many full circles the shaft makes. This concept is crucial since it's related to both the physics understanding and practical considerations in machinery, such as speed and distance coverage.