Problem 39 - University Physics with Modern Physics

Question

A hollow, thin-walled sphere of mass 12.0 kg and diameter 48.0 cm is rotating about an axle through its center. The angle (in radians) through which it turns as a function of time (in seconds) is given by \(\theta(t) = At^2 + Bt^4\), where \(A\) has numerical value 1.50 and \(B\) has numerical value 1.10. (a) What are the units of the constants \(A\) and \(B\)? (b) At the time 3.00 s, find (i) the angular momentum of the sphere and (ii) the net torque on the sphere.

Step-by-Step Solution

Answer

(a) Units: A is s^{-2}, B is s^{-4}. (b) (i) L = 58.9 kg·m^2/s, (ii) τ = 56.1 N·m.

Step 1: Determine Units of A and B

The angle \( \theta \) is given in radians, which is dimensionless. Therefore, the units on both sides of the equation \( \theta(t) = At^2 + Bt^4 \) must be consistent. For the term \( At^2 \), since the result \( \theta \) is dimensionless, the units of \( A \) must be \( s^{-2} \).For the term \( Bt^4 \), again, to remain dimensionless, the units of \( B \) must be \( s^{-4} \). Thus, the unit of \( A \) is \( s^{-2} \), and the unit of \( B \) is \( s^{-4} \).

Step 2: Calculate Moment of Inertia

For a hollow, thin-walled sphere, the moment of inertia \( I \) about its center is given by:\[I = \frac{2}{3} m r^2\]Here, \( m = 12.0 \) kg and the radius \( r = \frac{48.0}{2} \times 0.01 \) m = 0.24 m. Therefore, \[I = \frac{2}{3} \times 12.0 \times (0.24)^2 = 0.4608 \text{ kg} \cdot \text{m}^2\]

Step 3: Calculate Angular Velocity

The angular velocity \( \omega(t) \) is the derivative of \( \theta(t) \) with respect to time \( t \).\[\omega(t) = \frac{d}{dt}(At^2 + Bt^4) = 2At + 4Bt^3\]Substitute \( A = 1.50 \) and \( B = 1.10 \) to find \( \omega(3.00) \):\[\omega(3.00) = 2 \times 1.50 \times 3.00 + 4 \times 1.10 \times (3.00)^3 = 9.00 + 118.80 = 127.80 \text{ rad/s}\]

Step 4: Calculate Angular Momentum

The angular momentum \( L \) of the sphere is given by:\[L = I \times \omega\]Substituting the values \( I = 0.4608 \text{ kg} \cdot \text{m}^2 \) and \( \omega(3.00) = 127.80 \text{ rad/s} \):\[L = 0.4608 \times 127.80 = 58.9 \text{ kg} \cdot \text{m}^2/\text{s}\]

Step 5: Calculate Net Torque

The net torque \( \tau \) can be found as the derivative of angular momentum with respect to time or using \( \alpha = \frac{d\omega}{dt} \) where \( \tau = I \alpha \).\[\alpha(t) = \frac{d}{dt}(2At + 4Bt^3) = 2A + 12Bt^2\]Substituting \( t = 3.00 \) s, \( A = 1.50 \), and \( B = 1.10 \):\[\alpha(3.00) = 2 \times 1.50 + 12 \times 1.10 \times (3.00)^2 = 3.00 + 118.80 = 121.80 \text{ rad/s}^2\]Next, calculate the net torque:\[\tau = I \times \alpha = 0.4608 \times 121.80 = 56.1 \text{ N} \cdot \text{m}\]

Key Concepts

Angular Momentum

Angular momentum is a fundamental concept in rotational dynamics, similar to linear momentum in linear motion. It is a measure of the quantity of rotation of an object and is dependent on the rotational velocity and the distribution of mass around the object's axis of rotation. For the hollow sphere in the exercise, we calculate angular momentum, denoted as \( L \), using the formula:

• \( L = I \times \omega \)

where \( I \) is the moment of inertia, and \( \omega \) is the angular velocity. Angular momentum is an important value as it remains constant unless acted on by an external torque. This conservation helps in understanding rotational systems and predicting their future states when external forces are applied.

Moment of Inertia

Moment of inertia is a property that quantifies an object's resistance to changes in its rotational motion. In essence, it is like mass in linear motion but for rotation. For different shapes, moment of inertia is calculated differently; for a hollow sphere, it's given by:

• \( I = \frac{2}{3} m r^2 \)

where \( m \) is the mass, and \( r \) is the radius. This exercise illustrates calculating the moment of inertia as \( 0.4608 \, \text{kg} \cdot \text{m}^2 \). Understanding moment of inertia is crucial to predicting how force affects rotational motion, essentially dictating how difficult it is to change an object's angular velocity.

Angular Velocity

Angular velocity is a measure of how fast an object rotates or revolves relative to another point, often the center of its rotation. It's like how speed measures how fast something moves from point A to B in linear motion. Angular velocity, \( \omega \), is derived from the change in angle \( \theta \) over time \( t \). For the given sphere, angular velocity is calculated by:

• \( \omega(t) = \frac{d}{dt}(At^2 + Bt^4) = 2At + 4Bt^3 \)

Substitute the time to find \( \omega(3.00) = 127.80 \text{ rad/s} \). This value tells us how quickly the sphere is spinning, and is pivotal in calculating quantities like angular momentum.

Torque

Torque is a rotational force that causes an object to twist around an axis. It is analogous to force in linear dynamics but for rotation. Torque, \( \tau \), can change the angular momentum of a body, and is found by the relation:

• \( \tau = I \cdot \alpha \)

Here \( \alpha \) is the angular acceleration. Calculating torque helps to understand and predict how and why objects start, stop, or change their rotational motion. In the exercise, torque calculation gave us a value of \( 56.1 \, \text{N} \cdot \text{m} \). This value is essential in applications like engines and gears where rotational motion is key.

Angular Acceleration

Angular acceleration measures how quickly the angular velocity of an object changes with time. This is important in determining how rotational speeds alter over short or extended periods. It is given by:

• \( \alpha(t) = \frac{d}{dt}(2At + 4Bt^3) = 2A + 12Bt^2 \)

Substituting into the equation for this exercise produces \( \alpha(3.00) = 121.80 \text{ rad/s}^2 \). Angular acceleration tells us how much the speed of rotation is increasing or decreasing, providing insight into the dynamical behavior of rotating systems. This data is beneficial for designing mechanical systems like actuators, turbines, or any system where rotation is an element of function.