Converting Between Linear & Rotational: Videos & Practice Problems

Linear displacement (Δx) relates to angular displacement (Δθ) through the equation ${\mathrm{\Delta x}}^r=\mathrm{\Delta \theta }$. Tangential velocity (v_t) connects to angular velocity (ω) via $v^t=r\omega$. All points on a rigid body share the same angular quantities (Δθ, ω, α), while linear speeds vary with radial distance (r). Understanding these relationships is crucial for analyzing rotational motion and acceleration types, including tangential acceleration (a_t) and centripetal acceleration (a_c).

Converting Between Linear & Rotational

In the study of rotational motion, it's essential to understand the relationship between linear (or tangential) and angular (or rotational) quantities. Linear displacement, denoted as \(\Delta x\), is connected to angular displacement, \(\Delta \theta\), through the equation:

\[ \Delta x = r \Delta \theta \]

Here, \(r\) represents the radius of the circular path. Similarly, the relationship between tangential velocity (\(v_t\)) and angular velocity (\(\omega\)) is given by:

\[ v_t = r \omega \]

In this context, \(v_t\) refers to the linear velocity of a point on a rotating object, while \(\omega\) is the angular velocity. The pattern continues with acceleration, where tangential acceleration (\(a_t\)) relates to angular acceleration (\(\alpha\)) through the equation:

\[ a_t = r \alpha \]

Understanding these relationships is crucial when analyzing problems involving rotating bodies, such as a spinning disk. For instance, if a disk rotates with an angular speed of \(\omega = 10\) radians per second, we can determine the linear velocities at various points on the disk based on their distance from the center.

Consider a disk with a radius of 8 meters. If we analyze three points: one at the center (point 1), one at 4 meters from the center (point 2), and one at the edge (point 3), we find that:

1. All points share the same angular velocity, \(\omega\), which is 10 radians per second.

2. The linear velocities differ due to their varying distances from the center:

For point 1 (at the center):

\[ v_1 = r_1 \omega = 0 \times 10 = 0 \text{ m/s} \]

For point 2 (4 meters from the center):

\[ v_2 = r_2 \omega = 4 \times 10 = 40 \text{ m/s} \]

For point 3 (at the edge, 8 meters from the center):

\[ v_3 = r_3 \omega = 8 \times 10 = 80 \text{ m/s} \]

This illustrates that as the distance from the center increases, the linear velocity also increases. Thus, a point at the edge of the disk moves faster than a point closer to the center. This concept can be visualized by considering a carousel: the further you are from the center, the faster you move in a linear direction.

In summary, while all points on a rigid body in rotation share the same angular quantities (\(\Delta \theta\), \(\omega\), and \(\alpha\)), their linear velocities depend on their radial distances from the center, highlighting the distinction between angular and linear motion in rotational dynamics.

Length of string rotating point mass

In this scenario, we analyze the motion of a small object rotating at the end of a light string, forming a circular path. The object accelerates from rest to a final speed of 120 revolutions per minute (RPM) in 4 seconds. To understand the dynamics of this motion, we need to determine the length of the string, which corresponds to the radius of the circular path.

Initially, the object starts at rest, meaning its initial angular velocity (\( \omega_{\text{initial}} \)) is 0. The final angular velocity (\( \omega_{\text{final}} \)) can be calculated by converting RPM to radians per second. The conversion is done using the formula:

\( \omega_{\text{final}} = 2\pi f \)

where \( f \) is the frequency in revolutions per second. Since \( f = \frac{120 \text{ RPM}}{60} = 2 \text{ Hz} \), we find:

\( \omega_{\text{final}} = 2\pi \times 2 = 4\pi \text{ radians/second} \)

Next, we need to calculate the angular acceleration (\( \alpha \)). Using the equation of motion for angular velocity:

\( \omega_{\text{final}} = \omega_{\text{initial}} + \alpha \Delta t \)

Substituting the known values:

\( 4\pi = 0 + \alpha \times 4 \)

Solving for \( \alpha \) gives:

\( \alpha = \frac{4\pi}{4} = \pi \text{ radians/second}^2 \approx 3.14 \text{ radians/second}^2 \)

Now, we can relate the tangential acceleration (\( a_t \)) to the radius (\( r \)) and angular acceleration (\( \alpha \)) using the formula:

\( a_t = r \alpha \)

Given that the tangential acceleration is 15 meters per second squared, we can rearrange the equation to solve for \( r \):

\( r = \frac{a_t}{\alpha} = \frac{15}{\pi} \approx 4.77 \text{ meters}

This calculation reveals that the length of the string, or the radius of the circular path, is approximately 4.77 meters. This problem illustrates the interconnectedness of linear and angular motion, emphasizing the importance of understanding the relationships between different physical quantities in rotational dynamics.

A disc of radius 10 m rotates around itself with a constant 180 RPM. Calculate the linear speed at a point 7 m from the center of the disc.

• A. 18.8 m/s
• B. 21.0 m/s
• C. 66.0 m/s
• D. 132 m/s

Answer: (D)

A rock rotates around a light, 4-m long string. The rock is initially at rest, but reaches 150 RPM in 3 seconds. Calculate its tangential acceleration after 3 s. 
BONUS:Calculate its tangential speed after 3 s.

• A. 1.31 m/s²
• B. 15.8 m/s²
• C. 20.9 m/s²
• D. 62.7 m/s²

Answer: (C)

A 4 m long blade initially at rest begins to spin with 3 rad/s² around its axis, which is located at the middle of the blade. It accelerates for 10 s. Find the tangential speed of a point at the tip of the blade 10 s after it starts rotating.

• A. 0 m/s
• B. 15 m/s
• C. 30 m/s
• D. 60 m/s
• E. 188 m/s

Answer: (D)

Here’s what students ask on this topic:

How do you convert linear displacement to angular displacement?

Linear displacement (Δx) can be converted to angular displacement (Δθ) using the equation:

$\mathrm{\Delta x}=r\mathrm{\Delta \theta }$

Here, Δx is the linear displacement, r is the radius or distance from the center of rotation, and Δθ is the angular displacement. This relationship is crucial for understanding how a point on a rotating object moves in a straight line as the object rotates.

What is the relationship between tangential velocity and angular velocity?

Tangential velocity (v_t) is related to angular velocity (ω) through the equation:

$v$_t=rω

In this equation, v_t is the tangential velocity, r is the radius or distance from the center of rotation, and ω is the angular velocity. This relationship helps in determining the linear speed of a point on a rotating object.

How do you calculate tangential acceleration from angular acceleration?

Tangential acceleration (a_t) can be calculated from angular acceleration (α) using the equation:

$a$_t=rα

Here, a_t is the tangential acceleration, r is the radius or distance from the center of rotation, and α is the angular acceleration. This equation is useful for understanding how the linear acceleration of a point on a rotating object changes as the object accelerates angularly.

Why do all points on a rigid body have the same angular quantities?

All points on a rigid body have the same angular quantities (Δθ, ω, α) because they rotate together as a single unit. This means that the angular displacement (Δθ), angular velocity (ω), and angular acceleration (α) are uniform across the entire body. However, the linear speeds of different points vary with their radial distance (r) from the center of rotation. This uniformity in angular quantities is essential for analyzing rotational motion and ensuring consistent behavior across the rigid body.

How does radial distance affect linear speed in rotational motion?

In rotational motion, the linear speed (v) of a point on a rotating object depends on its radial distance (r) from the center of rotation. The relationship is given by:

$v=r\omega$

Here, v is the linear speed, r is the radial distance, and ω is the angular velocity. Points farther from the center (larger r) have higher linear speeds, while points closer to the center (smaller r) have lower linear speeds. This is why objects at the edge of a rotating disk move faster than those near the center.