Problem 24 - 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 \(0.250 \mathrm{rev} / \mathrm{s}\) and a constant angular acceleration of \(0.900 \mathrm{rev} / \mathrm{s}^{2}\). (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 \(t=0.200 \mathrm{~s} ?\) (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) 0.430 rev/s; (b) 0.068 revolutions; (c) 1.01 m/s; (d) 3.45 m/s²

Step 1: Understanding the Problem

We have a turntable of diameter 0.750 m rotating with initial angular velocity \( \omega_0 = 0.250 \, \mathrm{rev/s} \) and a constant angular acceleration \( \alpha = 0.900 \, \mathrm{rev/s^2} \). We need to solve for the angular velocity after 0.200 s, the number of revolutions in this time, tangential speed of a point on the rim, and the magnitude of resultant acceleration.

Step 1: Calculate the final angular velocity

Using the formula for angular velocity \( \omega = \omega_0 + \alpha \cdot t \), where \( t = 0.200 \, \mathrm{s} \), \( \omega_0 = 0.250 \, \mathrm{rev/s} \), and \( \alpha = 0.900 \, \mathrm{rev/s^2} \), we can compute the final angular velocity:\[ \omega = 0.250 + 0.900 \times 0.200 = 0.250 + 0.180 = 0.430 \, \mathrm{rev/s} \]

Step 2: Calculate the number of revolutions

The number of revolutions can be found using the formula for angular displacement \( \theta = \omega_0 \cdot t + 0.5 \cdot \alpha \cdot t^2 \):\[ \theta = 0.250 \times 0.200 + 0.5 \times 0.900 \times (0.200)^2 = 0.050 + 0.018 = 0.068 \, \mathrm{rev} \]

Step 3: Calculate the tangential speed

Tangential speed \( v_t \) is given by \( v_t = r \times \omega \), where the radius \( r = 0.750/2 = 0.375 \, \mathrm{m} \) and \( \omega = 0.430 \, \mathrm{rev/s} = 0.430 \times 2\pi \, \mathrm{rad/s} \):\[ \omega = 0.430 \times 2\pi = 2.701 \approx 2.70\, \mathrm{rad/s} \]\[ v_t = 0.375 \times 2.70 \approx 1.01 \, \mathrm{m/s} \]

Step 4: Calculate the resultant acceleration

The resultant acceleration consists of both tangential \( a_t \) and radial (centripetal) \( a_c \) components.For tangential acceleration, use \( a_t = r \cdot \alpha \) with \( \alpha = 0.900 \times 2\pi \):\[ a_t = 0.375 \times 0.900 \times 2\pi = 2.12 \approx 2.12 \, \mathrm{m/s^2} \]Centripetal acceleration \( a_c \) is \( a_c = r \cdot \omega^2 \):\[ a_c = 0.375 \times (2.70)^2 \approx 2.73 \, \mathrm{m/s^2} \]Resultant acceleration \( a = \sqrt{a_t^2 + a_c^2} \):\[ a = \sqrt{2.12^2 + 2.73^2} \approx 3.45 \approx 3.45 \, \mathrm{m/s^2} \]

Key Concepts

Tangential Speed

Understanding tangential speed is essential when dealing with rotational motion. When an object is moving in a circular path, tangential speed refers to how fast the object is traveling along that path at any given point. It's the linear speed of a point on the circumference of a rotating object.

Calculating Tangential Speed:

• The formula for tangential speed (\( v_t \)) is: \( v_t = r \cdot \omega \). Here, \( r \) is the radius of the path, and \( \omega \) is the angular velocity of the rotating object.
• In the given exercise, with the diameter of 0.750 m, the radius is half of that, 0.375 m. The angular velocity at \( t = 0.200 \) seconds is computed as 0.430 rev/s, which needs conversion to rad/s to use in calculations because the equation for tangential speed uses SI units.
• Converting the angular velocity: \( 0.430 \times 2\pi \approx 2.70 \, \mathrm{rad/s} \).
• Now, the tangential speed: \( 0.375 \times 2.70 \approx 1.01 \, \mathrm{m/s} \).

In essence, tangential speed tells us how quickly a point on the rim of a spinning disk would move if it were suddenly set free to travel in a straight line.

Angular Velocity

Angular velocity is a measure of how much an object spins around a specific axis over a certain amount of time. It answers the question of how fast the angle is changing in a rotational motion.

Comprehending Angular Velocity:

• Represented by \( \omega \), angular velocity is the rate of change of the angle with which an object revolves around a circle or an axis.
• Its unit is typically radians per second (rad/s), though it may also be expressed in revolutions per second (rev/s), as seen in the exercise where \( \omega \) initially is 0.250 rev/s.
• To calculate the final angular velocity given constant angular acceleration, use: \( \omega = \omega_0 + \alpha \cdot t \).
• In the exercise, applying this formula: \( \omega = 0.250 + 0.900 \times 0.200 = 0.430 \, \text{rev/s} \).

A key point to remember about angular velocity is that it's directly related to both the speed at which an object rotates and the radius of its path.

Centripetal Acceleration

Centripetal acceleration is crucial in rotational dynamics as it describes the acceleration required to keep an object moving in a circular path, constantly pulling it towards the center of the circle.

Explaining Centripetal Acceleration:

• It maintains an object's circular motion and is defined by \( a_c = r \cdot \omega^2 \), where \( \omega \) is the angular velocity and \( r \) is the radius of the circle.
• In our problem, \( \omega = 2.70 \, \mathrm{rad/s} \) and \( r = 0.375 \, \mathrm{m} \), allowing us to calculate \( a_c \) as: \( a_c = 0.375 \times (2.70)^2 \approx 2.73 \, \mathrm{m/s^2} \).
• Centripetal acceleration is always directed towards the center of the circular path, hence "centripetal," meaning "center-seeking."

This type of acceleration is essential for understanding how forces act on objects moving in circular paths. It shows that without centripetal acceleration, objects would move off in a tangent straight line due to inertia.