Question

A top spins for 10.0 min, beginning with an angular speed of 10.0 rev/s. Determine its angular acceleration, assuming it is constant, and its total angular displacement.

Short answer

Answer: The angular acceleration (α) of the top is approximately -0.000183 rad/s², and the total angular displacement (θ) during the top's spinning is approximately -2.16 x 10⁷ radians.

Step-by-step solution

1. Convert the given time to seconds

The given time is 10.0 minutes. Convert this to seconds by multiplying with 60s (1 min = 60s). Time (t) = 10.0 min × 60s = 600s

2. Write down the given information and symbols

Let's use the following symbols for the given information: Initial angular speed (ω₀) = 10.0 rev/s Angular acceleration (α) = ? Time (t) = 600s Angular displacement (θ) = ?

3. Find the final angular speed (ω) using the given information and the angular motion formula

Since we know the initial angular speed, the time, and we are asked to find the angular acceleration, we can use the following angular motion formula: ω = ω₀ + αt In order to find α, we first need to find ω (final angular speed) for which we will use the formula: ω² = ω₀² + 2αθ

4. Calculate the final angular speed (ω) using the given information

Before we can use the formula ω² = ω₀² + 2αθ, we need to convert the initial angular speed from rev/s to rad/s to keep everything in SI units. Recall that 1 revolution = 2π radians. ω₀ = 10.0 rev/s × (2π rad/rev) = 20π rad/s Now, we can rewrite the formula as: ω = sqrt(ω₀² + 2αθ) In this case, the top stops spinning, which means the final angular speed is 0 rad/s: 0 = sqrt((20π)² + 2αθ)

5. Determine the angular acceleration (α) from the formula

Now, we need to rewrite the previous formula to isolate α and solve for it: 0 = (20π)² + 2αθ - (20π)² = 2αθ α = -(20π)² / (2θ) Since we need θ as well, let's solve for it in another angular motion formula: θ = ω₀t + 0.5αt²

6. Solve for the angular displacement (θ) in the equation

We can substitute values into the formula and solve for θ: θ = (20π)(600) + 0.5α(600)² We already have an expression for α, so substitute that back into the equation: θ= (20π)(600) + 0.5(- (20π)² / (2θ))(600)² θ = 12000π + (-(20π)²)(600)² / (2θ) Now, solve for θ numerically: θ ≈ -2.16 x 10⁷ rad

7. Calculate the angular acceleration (α) using the angular displacement (θ)

Using the previously derived expression for α, we can plug in our values for θ: α = -(20π)² / (2(-2.16 x 10⁷)) α ≈ -0.000183 rad/s²

8. Summarize the results

The angular acceleration (α) of the top is approximately -0.000183 rad/s². This means that the top is decelerating at a constant rate. The total angular displacement (θ) during the top's spinning is approximately -2.16 x 10⁷ radians.

Key concepts

Angular Speed

Angular speed refers to how fast an object rotates or revolves around a fixed point or axis. In physics, it's an essential concept when examining the motion of anything that spins, from a simple wheel to celestial bodies. It's given the symbol \( \omega \) and can be expressed in units such as revolutions per minute (rpm) or, in the International System of Units (SI), radians per second (rad/s).

To understand angular speed in the context of the exercise, consider a top spinning at an initial angular speed of 10.0 revolutions per second. To use equations of motion that are standardized in physics, this speed must be converted into radians per second using the conversion factor \( 1 \text{ rev} = 2\pi \text{ rad} \). Therefore, the initial angular speed \( \omega_0 \) is \( 10.0 \text{ rev/s} \times 2\pi \text{ rad/rev} = 20\pi \text{ rad/s} \).

The concept of angular speed is vital because it represents the rate at which the angular position of the spinning object changes over time. When the angular speed decreases, this indicates that the object is experiencing a deceleration, and when it increases, the object is accelerating.

Angular Displacement

Angular displacement is the measure of the angle through which an object has rotated in a particular time. It's denoted by \( \theta \) and describes the object's change in angular position. In our spinning top scenario, the angular displacement would be the total angle the top has turned through during its 10 minutes of spinning.

Angular displacement, like angular speed, is usually measured in radians (where \( 2\pi \text{ rad} \) equals one full revolution). The solution to the exercise involved a calculation where the top's angular displacement, given enough information, can be determined by the formula \( \theta = \omega_0t + 0.5\alpha t^2 \), taking the initial angular speed (\( \omega_0 \) ) and the constant angular acceleration (\( \alpha \) ) into account. After a complex series of computations, the angular displacement comes out to be approximately -2.16 x 10⁷ radians, indicating the top has rotated through that angle in the negative direction due to the deceleration.

Angular Motion Formulas

Angular motion formulas are equations used to describe the motion of rotating objects. They are similar to the linear motion equations but are used for rotational dynamics instead. Key angular motion formulas include \( \omega = \omega_0 + \alpha t \) for the final angular speed, and \( \theta = \omega_0t + 0.5\alpha t^2 \) to find the angular displacement.

In the textbook solution, these formulas were employed to calculate the angular acceleration \( \alpha \) and the angular displacement \( \theta \) of the spinning top. The process required substituting known values and solving for unknowns. For instance, we can determine the angular acceleration by rearranging the equation \( \omega^2 = \omega_0^2 + 2\alpha\theta \), which resulted in finding that the top's angular acceleration was approximately -0.000183 rad/s². This value illustrates that the top was slowing down over time at a constant rate.

The angular motion formulas are critical tools for understanding rotational motion. When they're used correctly, these equations allow us to predict future motion, which is essential in many fields ranging from mechanical engineering to astrophysics. Moreover, understanding the principles behind these equations enables students to tackle a myriad of problems dealing with rotation in a methodical and logical manner.