Problem 56 - University Physics with Modern Physics

Question

A uniform disk with radius \(R =\) 0.400 m and mass 30.0 kg rotates in a horizontal plane on a frictionless vertical axle that passes through the center of the disk. The angle through which the disk has turned varies with time according to \(\theta(t) =\) (1.10 rad/s)\(t +\) (6.30 rad/s\(^2)t^2\). What is the resultant linear acceleration of a point on the rim of the disk at the instant when the disk has turned through 0.100 rev?

Step-by-Step Solution

Answer

Resulting linear acceleration is calculated by combining tangential and centripetal components using vector addition.

Step 1: Convert Revolutions to Radians

The first task is to convert 0.100 revolutions into radians, since angular equations are typically in radians. Since one full revolution is \(2\pi\) radians:\[0.100 \text{ rev} \times 2\pi \text{ rad/rev} = 0.200\pi \text{ rad}\].

Step 2: Solve for Time when \(\theta(t)\) equals 0.200\pi rad

Set \(\theta(t) = 0.200\pi \text{ rad}\) in the given equation \(\theta(t) = (1.10 \text{ rad/s})t + (6.30 \text{ rad/s}^2)t^2\) and solve for \(t\).\[0.200 \pi = 1.10t + 6.30t^2\]Rearrange to form a quadratic equation:\[6.30t^2 + 1.10t - 0.200\pi = 0\]Use the quadratic formula, \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), to find \(t\).Here, \(a = 6.30\), \(b = 1.10\), and \(c = -0.200\pi\).

Step 3: Calculate the Angular Velocity \(\omega(t)\)

Find \(\omega(t)\) by taking the derivative of \(\theta(t)\), which gives the angular velocity:\[\omega(t) = \frac{d\theta(t)}{dt} = 1.10 + 2 \times 6.30t = 1.10 + 12.60t\].Substitute the value of \(t\) found from Step 2 to find \(\omega\).

Step 4: Determine Angular Acceleration \(\alpha\)

Angular acceleration \(\alpha\) is the second derivative of \(\theta(t)\), giving:\[\alpha(t) = \frac{d^2\theta(t)}{dt^2} = 12.60 \text{ rad/s}^2\].The constant term indicates that angular acceleration remains constant at all times.

Step 5: Find the Tangential Acceleration \(a_t\)

Tangential acceleration is given by \(a_t = R \times \alpha\):\[a_t = 0.400 \times 12.60 = 5.04 \text{ m/s}^2\].

Step 6: Compute the Centripetal Acceleration \(a_c\)

Centripetal acceleration is given by \(a_c = R \times \omega^2\):First calculate \(\omega\) using the value of \(t\) found in Step 3, then:\[a_c = 0.400 \times \omega^2\].

Step 7: Calculate the Resultant Linear Acceleration

The resultant linear acceleration of the point on the rim of the disk is obtained by combining tangential and centripetal accelerations:\[a = \sqrt{a_t^2 + a_c^2}\].This is a vector sum of the two, as they are perpendicular to each other.

Key Concepts

Angular Velocity

Angular velocity refers to the rate at which an object rotates around an axis. It's a measure of how quickly the angle changes over time. The symbol for angular velocity is \( \omega \), and it's typically expressed in radians per second (rad/s).

In rotational dynamics, angular velocity is analogous to linear velocity in linear motion. It tells you how fast a point on the rotating object is moving in a circular path. For our given problem, the angular velocity is derived from the equation \(\theta(t) = (1.10 \text{ rad/s})t + (6.30 \text{ rad/s}^2)t^2\).

To find \(\omega(t)\), we take the derivative of \(\theta(t)\) with respect to time \(t\), yielding:

• \(\omega(t) = \frac{d\theta(t)}{dt} = 1.10 + 12.60t\)

This formula tells us that the angular velocity is dependent on time and includes a constant term (1.10 rad/s) and a time-dependent term (12.60t), reflecting a typical scenario in rotational motion with constant angular acceleration.

Angular Acceleration

Angular acceleration is the rate of change of angular velocity over time, representing how quickly an object speeds up or slows down its rotation. Denoted by \( \alpha \), angular acceleration is measured in radians per second squared (rad/s²).

In the context of the exercise, the angular acceleration \(\alpha\) is obtained by taking the second derivative of the angular position function \(\theta(t)\):

• \(\alpha(t) = \frac{d^2\theta(t)}{dt^2} = 12.60 \text{ rad/s}^2\)

This indicates a constant angular acceleration, meaning the rate of change of angular velocity is steady. Constant angular acceleration simplifies calculations significantly since it allows us to use simplified kinematic equations to analyze rotational motion, just like constant linear acceleration does in linear dynamics.

Centripetal Acceleration

Centripetal acceleration plays a vital role in circular motion, acting towards the center of rotation to keep an object moving along a curved path. It ensures a changing direction while maintaining the circular motion. Calculated using the formula \( a_c = R \times \omega^2 \), centripetal acceleration is measured in meters per second squared (m/s²).

For this exercise, knowing \(R = 0.400\) m and the value of \(\omega\) from previous calculations, you can determine the centripetal acceleration:

• First, calculate \(\omega\) using \(\omega(t) = 1.10 + 12.60t\)
• Then compute \( a_c \) using \( a_c = R \times \omega^2 \)

Centripetal acceleration depends on both the radius of the circular path and the square of the angular velocity, demonstrating that a faster rotation or a larger radius results in greater centripetal force required to maintain the circular motion.

Tangential Acceleration

Tangential acceleration in rotational dynamics is analogous to linear acceleration, referring to the rate at which the speed of a point along a circular path changes. While centripetal acceleration changes the direction, tangential acceleration changes the speed of rotation.

Defined as \( a_t = R \times \alpha \), it highlights how angular acceleration directly influences the linear acceleration of points along the rotating body. In the given problem, here's how to find \( a_t \):

• Use \( R = 0.400 \text{ m} \) and \( \alpha = 12.60 \text{ rad/s}^2 \)
• Calculate \( a_t = 0.400 \times 12.60 = 5.04 \text{ m/s}^2 \)

Combining tangential and centripetal accelerations gives a comprehensive view of how a point on the disk moves. Tangential acceleration affects the rotational speed, essential for understanding changes in rotational motion over time.