Problem 5 - University Physics with Modern Physics

Question

A child is pushing a merry-go-round. The angle through which the merry-go- round has turned varies with time according to \(\theta(t)=\gamma t+\beta t^{3},\) where \(\gamma=0.400 \mathrm{rad} / \mathrm{s}\) and \(\beta=0.0120 \mathrm{rad} / \mathrm{s}^{3}\) . (a) Calculate the angular velocity of the merry-go-round as a function of time. (b) What is the initial value of the angular velocity? (c) Calculate the instantaneous value of the angular velocity \(\omega_{z}\) at \(t=5.00 \mathrm{s}\) and the average angular velocity \(\omega_{\mathrm{av}-\mathrm{z}}\) for the time interval \(t=0\) to \(t=5.00 \mathrm{s}\) . Show that \(\omega_{\mathrm{av}-\mathrm{z}}\) is not equal to the average of the instantaneous angular velocities at \(t=0\) and \(t=5.00 \mathrm{s},\) and explain why it is not.

Step-by-Step Solution

Answer

Instantaneous angular velocity at 5.00s is 1.30 rad/s; the average is 0.700 rad/s, different from the midpoint average of 0.850 rad/s.

Step 1: Identify the formula for angular velocity

The angular velocity \(\omega(t)\) is the derivative of the angular displacement \(\theta(t)\) with respect to time \(t\). Given the equation \(\theta(t) = \gamma t + \beta t^3\), the formula for angular velocity becomes: \[ \omega(t) = \frac{d\theta}{dt} = \frac{d}{dt}(\gamma t + \beta t^3). \]

Step 2: Calculate the derivative

Apply differentiation to each term: 1. The derivative of \(\gamma t\) with respect to \(t\) is \(\gamma\).2. The derivative of \(\beta t^3\) with respect to \(t\) is \(3\beta t^2\).Combine these results to get: \[ \omega(t) = \gamma + 3\beta t^2. \]

Step 3: Substitute values for angular velocity function

Substitute the given values of \(\gamma = 0.400\, \mathrm{rad/s}\) and \(\beta = 0.0120\, \mathrm{rad/s^3}\) into the angular velocity equation: \[ \omega(t) = 0.400 + 3 \times 0.0120 \times t^2 = 0.400 + 0.036 t^2. \]

Step 4: Calculate initial angular velocity

The initial angular velocity is the value of \(\omega(t)\) when \(t = 0\). \[ \omega(0) = 0.400 + 0.036 \times 0^2 = 0.400 \mathrm{rad/s}. \]

Step 5: Calculate instantaneous angular velocity at t=5.00s

Substitute \(t = 5.00\, \mathrm{s}\) into the angular velocity function:\[ \omega(5.00) = 0.400 + 0.036 \times (5.00)^2 = 0.400 + 0.036 \times 25.00 = 1.30 \mathrm{rad/s}. \]

Step 6: Calculate average angular velocity from t=0 to t=5.00s

The average angular velocity \(\omega_{\mathrm{av}-z}\) over a time interval \(\Delta t\) is: \[ \omega_{\mathrm{av}-z} = \frac{\theta(5.00) - \theta(0)}{5.00 - 0}. \]First, calculate \(\theta(5.00)\) and \(\theta(0)\): \[ \theta(5.00) = 0.400 \times 5.00 + 0.0120 \times (5.00)^3 = 2.00 + 1.50 = 3.50 \, \mathrm{rad}, \] \[ \theta(0) = 0. \]Thus, the average angular velocity is: \[ \omega_{\mathrm{av}-z} = \frac{3.50 - 0}{5.00} = 0.700 \, \mathrm{rad/s}. \]

Step 7: Compare to the average of instantaneous velocities

The average of the instantaneous angular velocities at \(t=0\) and \(t=5.00\) is: \[ \frac{\omega(0) + \omega(5.00)}{2} = \frac{0.400 + 1.30}{2} = 0.850 \, \mathrm{rad/s}. \]Notice that \(\omega_{\mathrm{av}-z} = 0.700 \, \mathrm{rad/s}\) is not equal to \(0.850 \, \mathrm{rad/s}\). This is because the angular velocity is not constant over the interval, due to the \(t^2\) term in its expression, causing the rate of change of angular displacement to be non-linear.

Key Concepts

Merry-go-round Dynamics

When thinking about merry-go-round dynamics, imagine the joyful circular motion of children at a playground. Despite its simplicity, the mechanics behind a merry-go-round involve interesting physics, particularly the concept of angular velocity which measures how quickly it spins. Angular velocity, denoted as \(\omega\), indicates how fast something rotates per unit time.In the given problem, the merry-go-round's turning is described by the function \(\theta(t) = \gamma t + \beta t^3\), where \(\theta(t)\) indicates the angular position. It is made up of two parts that contribute to its dynamics:

• \(\gamma t\) : Represents a constant spin. It dictates the uniform angular motion.
• \(\beta t^3\) : Introduces variability, accounting for acceleration during the spin.

The combination of these factors determines how the merry-go-round speeds up or maintains its pace over time.

Time-dependent Functions in Physics

In the realm of physics, phenomena are often expressed through time-dependent functions which model how things change. These functions, like \(\theta(t)\) in the problem, describe the system's behavior over time. Here, they are crucial in predicting the merry-go-round’s dynamics.The function \(\theta(t) = \gamma t + \beta t^3\) captures both the regular motion and the changing acceleration. In real-world scenarios, several factors influence how these functions look:

• Constants (\(\gamma\)) often represent factors that remain unchanged over time.
• Variable components (\(\beta t^3\)) introduce time-based changes, often linked to acceleration or deceleration.

These kinds of functions are indispensable, as they allow us to extract vital information like the instantaneous state of motion and predict future dynamics.

Differentiation in Physics

Differentiation, one of the most powerful mathematical tools in physics, allows us to explore how quantities change over time. Just like how a speedometer tells us how fast a car is moving at a specific moment, differentiation helps us find the rate at which a position changes — essentially, finding angular velocity from a position function.In the merry-go-round problem, differentiation is used to determine \(\omega(t)\) from \(\theta(t)\). By applying the derivative operation:

• For \(\gamma t\), the constant \(\gamma\) remains because it reflects a steady rate.
• For \(\beta t^3\), it becomes \(3\beta t^2\), showing how the rate changes with time.

Hence, the angular velocity as a function is: \(\omega(t) = \gamma + 3\beta t^2\). This equation highlights the power of differentiation in predicting dynamic motion, illustrating how rotational speeds accelerate or maintain based on time.