Torque & Acceleration (Rotational Dynamics): Videos & Practice Problems

Torque causes angular acceleration, similar to how force causes linear acceleration. The relationship is expressed as $\Sigma \left(\tau \right)=I\alpha$ for rotational dynamics, where $I$ is the moment of inertia and $\alpha$ is angular acceleration. For point masses, the moment of inertia is $m\cdot r^2$. Understanding these concepts is crucial for solving problems involving rotational motion and dynamics.

Torque & Acceleration (Rotational Dynamics)

In the study of rotational dynamics, we explore the relationship between torque and angular acceleration. Torque, which can be thought of as the rotational equivalent of force, causes an object to accelerate rotationally. This concept is crucial in understanding how forces lead to motion in a circular path.

When a force is applied to an object, it produces a torque, which is defined mathematically as:

$$\tau = r \cdot F \cdot \sin(\theta)$$

where:

• $$\tau$$ is the torque,
• $$r$$ is the distance from the axis of rotation to the point where the force is applied,
• $$F$$ is the applied force, and
• $$\theta$$ is the angle between the force vector and the radius vector.

In rotational dynamics, the relationship between torque and angular acceleration ($$\alpha$$) is analogous to Newton's second law for linear motion, which states that the sum of forces equals mass times acceleration ($$F = ma$$). The rotational equivalent is expressed as:

$$\sum \tau = I \cdot \alpha$$

where:

• $$\sum \tau$$ is the sum of all torques acting on the object,
• $$I$$ is the moment of inertia, which represents the object's resistance to angular acceleration, and
• $$\alpha$$ is the angular acceleration.

For a solid disc, the moment of inertia is given by:

$$I = \frac{1}{2} m r^2$$

where $$m$$ is the mass and $$r$$ is the radius of the disc. This relationship highlights that the distribution of mass relative to the axis of rotation significantly affects the rotational motion.

To find the angular acceleration resulting from a torque, one can rearrange the equation to solve for $$\alpha$$:

$$\alpha = \frac{\sum \tau}{I}$$

For example, consider a solid disc with a mass of 100 kg and a radius of 2 m, which is subjected to a tangential force of 50 N. The torque produced by this force can be calculated, and using the moment of inertia, we can derive the expression for angular acceleration. The torque due to the applied force is:

$$\tau = r \cdot F \cdot \sin(90^\circ) = r \cdot F$$

Substituting the values, we find:

$$\tau = 2 \cdot 50 = 100 \, \text{N m}$$

Now, substituting this torque into the equation for angular acceleration gives:

$$\alpha = \frac{100}{\frac{1}{2} \cdot 100 \cdot 2^2} = \frac{100}{200} = 0.5 \, \text{radians per second squared}$$

This example illustrates how to connect the concepts of torque, moment of inertia, and angular acceleration, reinforcing the principles of rotational dynamics. Understanding these relationships is essential for analyzing systems involving rotational motion.

Suppose that piano has a long, thin bar ran through it (totally random), shown below as the vertical red line, so that it is free to rotate about a vertical axis through the bar. You push the piano with a horizontal 100 N (blue arrow), causing it to spin about its vertical axis with 0.3 rad/s² . Your force acts at a distance of 1.1 m from the bar, and is perpendicular to a line connecting it to the bar (green dotted line). What is the piano's moment of inertia about its vertical axis?

• A. 250 kg•m²
• B. 285 kg•m²
• C. 303 kg•m²
• D. 367 kg•m²

Answer: (D)

Torque & Acceleration of a Point Mass

In the study of rotational dynamics, understanding how torque produces acceleration on a point mass is essential. A point mass is defined as a tiny object with no radius or volume, which simplifies the analysis of torque and acceleration. While most torque problems involve rigid bodies, examining point masses provides valuable insights into the underlying principles.

Consider a practical example: a small rock with a mass of 2 kilograms is spun at the end of a light string that is 3 meters long. In this scenario, the length of the string represents the distance from the center of rotation, denoted as \( r \). Therefore, \( r = 3 \, \text{m} \). To determine the net torque required to achieve an angular acceleration (\( \alpha \)) of 4 radians per second squared, we can use the relationship between net torque and angular acceleration, expressed as:

\[ \text{Net Torque} = \sum \tau = I \alpha \]

Here, \( I \) represents the moment of inertia. For point masses, the moment of inertia is calculated using the formula:

\[ I = m r^2 \]

Substituting the known values, we have:

\[ I = 2 \, \text{kg} \times (3 \, \text{m})^2 = 18 \, \text{kg} \cdot \text{m}^2 \]

Now, substituting \( I \) and \( \alpha \) into the torque equation gives:

\[ \text{Net Torque} = 18 \, \text{kg} \cdot \text{m}^2 \times 4 \, \text{rad/s}^2 = 72 \, \text{N} \cdot \text{m} \]

This result indicates that a net torque of 72 Newton meters is required to achieve the desired angular acceleration.

Next, to find the tangential acceleration (\( a_{\text{tan}} \)), we can use the relationship between tangential acceleration and angular acceleration, given by:

\[ a_{\text{tan}} = r \alpha \]

Substituting the values of \( r \) and \( \alpha \) yields:

\[ a_{\text{tan}} = 3 \, \text{m} \times 4 \, \text{rad/s}^2 = 12 \, \text{m/s}^2 \]

This indicates that the tangential acceleration at that point is 12 meters per second squared, representing a linear acceleration. Understanding these relationships between torque, angular acceleration, and tangential acceleration is crucial for solving problems in rotational dynamics.

Here’s what students ask on this topic:

What is the relationship between torque and angular acceleration?

The relationship between torque (τ) and angular acceleration (α) is given by the equation:

$\sum \tau =I\alpha$

Here, $I$ is the moment of inertia, which measures an object's resistance to changes in its rotational motion. This equation is analogous to Newton's second law for linear motion, $\sum F=ma$, where force causes linear acceleration. In rotational dynamics, torque causes angular acceleration.

How do you calculate the moment of inertia for a point mass?

The moment of inertia (I) for a point mass is calculated using the formula:

$I=m\cdot r^2$

Here, $m$ is the mass of the point object, and $r$ is the distance from the point mass to the axis of rotation. This formula is essential for solving problems involving rotational motion of point masses.

What is the difference between linear dynamics and rotational dynamics?

Linear dynamics deals with forces causing linear acceleration, described by Newton's second law:

$\sum F=ma$

where $F$ is the force, $m$ is the mass, and $a$ is the linear acceleration. Rotational dynamics, on the other hand, involves torques causing angular acceleration, described by:

$\sum \tau =I\alpha$

where $\tau$ is the torque, $I$ is the moment of inertia, and $\alpha$ is the angular acceleration.

How do you calculate the angular acceleration of a rotating object?

To calculate the angular acceleration (α) of a rotating object, you use the equation:

$\sum \tau =I\alpha$

First, determine the net torque (τ) acting on the object and its moment of inertia (I). Then, solve for α:

$\alpha =\frac{\sum }{\tau }$

This will give you the angular acceleration in radians per second squared (rad/s²).

What is the significance of the moment of inertia in rotational dynamics?

The moment of inertia (I) is a measure of an object's resistance to changes in its rotational motion. It plays a role similar to mass in linear dynamics. The larger the moment of inertia, the more torque is required to achieve the same angular acceleration. It depends on the mass distribution relative to the axis of rotation. For example, a solid disc has a different moment of inertia compared to a point mass or a hollow cylinder, even if they have the same mass.