Problem 26 - University Physics with Modern Physics

Question

An electric turntable 0.750 \(\mathrm{m}\) in diameter is rotating about a fixed axis with an initial angular velocity of 0.250 rev/s and a constant angular acceleration of 0.900 rev \(/ \mathrm{s}^{2} .\) (a) Compute the angular velocity of the turntable after 0.200 \(\mathrm{s}\) . (b) Through how many revolutions has the turntable spun in this time interval? (c) What is the tangential speed of a point on the rim of the turntable at \(t=0.200 \mathrm{s} ?(\mathrm{d})\) What is the magnitude of the resultant acceleration of a point on the rim at \(t=0.200 \mathrm{s} ?\)

Step-by-Step Solution

Answer

(a) \(0.86\pi\) rad/s, (b) 0.068 revolutions, (c) \(1.012\) m/s, (d) \(3.45\) m/s².

Step 1: Convert Units for Angular Velocity and Acceleration

First, ensure that the initial angular velocity \( \omega_0 \) and the angular acceleration \( \alpha \) are in consistent units. We have \( \omega_0 = 0.250 \) rev/s \( = 0.250 \times 2\pi = 0.5\pi \) rad/s and \( \alpha = 0.900 \) rev/s² \( = 0.900 \times 2\pi = 1.8\pi \) rad/s².

Step 2: Calculate Angular Velocity after 0.200 s

Use the equation for angular velocity under constant acceleration: \( \omega = \omega_0 + \alpha t \). Substitute the known values: \( \omega = 0.5\pi + 1.8\pi \times 0.200 \). Simplifying, \( \omega = 0.5\pi + 0.36\pi = 0.86\pi \).

Step 3: Calculate Number of Revolutions in 0.200 s

Use the equation for angular displacement \( \theta = \omega_0 t + 0.5 \alpha t^2 \). Substitute the known values: \( \theta = 0.5\pi \times 0.200 + 0.5 \times 1.8\pi \times (0.200)^2 \). Simplifying, \( \theta = 0.1\pi + 0.036\pi = 0.136\pi \) rad. Convert this to revolutions: \( \theta = \frac{0.136\pi}{2\pi} = 0.068 \) revolutions.

Step 4: Calculate Tangential Speed at the Rim

Tangential speed \( v \) at the rim is given by \( v = r\omega \), where \( r \) is the radius of the turntable. The radius \( r = 0.750/2 = 0.375 \) m. Substitute \( \omega = 0.86\pi \): \( v = 0.375 \times 0.86\pi \approx 1.012 \) m/s.

Step 5: Calculate Resultant Acceleration at the Rim

Calculate the tangential acceleration \( a_t = r\alpha \). Substitute \( r = 0.375 \) m and \( \alpha = 1.8\pi \): \( a_t = 0.375 \times 1.8\pi \approx 2.12 \) m/s². The centripetal acceleration \( a_c = \omega^2 r \). Substitute \( \omega = 0.86\pi \): \( a_c = (0.86\pi)^2 \times 0.375 \approx 2.73 \) m/s². The total acceleration \( a_r = \sqrt{a_t^2 + a_c^2} \approx \sqrt{2.12^2 + 2.73^2} \approx 3.45 \) m/s².

Key Concepts

Angular Velocity

Angular velocity refers to how quickly an object rotates around a central point or axis. It’s measured in radians per second (rad/s). When discussing a turntable, we often begin with its initial angular velocity. In the exercise, the turntable starts with an angular velocity of 0.250 revolutions per second. However, we convert this to radians per second to maintain consistency in calculations. This value is multiplied by \(2\pi\) because one revolution is equal to \(2\pi\) radians. Hence, initially, the angular velocity is \(0.5\pi\) rad/s. Over a given time period, an object under constant angular acceleration will see its angular velocity increase according to the formula:

• \( \omega = \omega_0 + \alpha t \)

Here, \(\omega_0\) is the initial angular velocity, \(\alpha\) is the angular acceleration, and \(t\) is the time elapsed. Calculating this for a period of 0.200 seconds in the exercise, results in a final angular velocity of \(0.86\pi\) rad/s for the turntable after 0.200 seconds under constant acceleration.

Angular Acceleration

Angular acceleration is the rate at which the angular velocity of an object changes with time. It’s an essential aspect of rotational motion, telling us how quickly an object is speeding up or slowing down its rotation. The unit for angular acceleration is radians per second squared (rad/s²). For the turntable exercise, the given angular acceleration is 0.900 rev/s², which converts to \(1.8\pi\) rad/s². Angular acceleration helps in determining how the speed of rotation increases over a particular period. Using the equation:

• \( \theta = \omega_0 t + 0.5 \alpha t^2 \)

we can calculate the number of revolutions an object makes over a certain time. Here, \(\theta\) represents the angular displacement in radians, showing how far the object has rotated. In 0.200 seconds, the turntable completes approximately 0.068 revolutions, effectively illustrating the concept of angular acceleration.

Tangential Speed

Tangential speed describes the linear speed of any point located along the outer edge of a circular path. It’s different from angular velocity, which considers rotational speed around the axis. In the case of the rotating turntable, tangential speed is given by the formula:

• \( v = r\omega \)

where \(r\) is the radius of the circular path and \(\omega\) is the angular velocity. For points on the edge of the turntable, this speed tells us how fast they are moving in a straight line as they rotate around the axis. Given the diameter of 0.750 m for the turntable, the radius \(r\) is half of this, equalling 0.375 m. With an angular velocity of \(0.86\pi\) rad/s, the tangential speed becomes approximately 1.012 m/s. This value represents how quickly a point on the rim is moving along its circular path, emphasizing the interplay between linear and angular movement.

Centripetal Acceleration

Centripetal acceleration is vital for any object moving in a circular path, ensuring that it remains on its path without flying outwards. It is directed towards the center of the circle and can be calculated with:

• \( a_c = \omega^2 r \)

where \(\omega\) is the angular velocity and \(r\) is the radius of the circle. In the context of the turntable exercise, with a radius of 0.375 m and an angular velocity of \(0.86\pi\) rad/s, the centripetal acceleration is calculated to be approximately 2.73 m/s².This acceleration ensures that a point on the rim does not deviate from its circular path. Combined with tangential acceleration, the resultant acceleration can provide a full picture of the motion by considering both changes in rotational speed and the constant adjustment needed to maintain the circular path. In this exercise, the total resultant acceleration at 0.200 seconds is found to be 3.45 m/s². This kind of comprehensive understanding is crucial for advanced studies involving rotational dynamics.