Question

A point on a Blu-ray disc is a distance \(R / 4\) from the axis of rotation. How far from the axis of rotation is a second point that has, at any instant, a linear velocity twice that of the first point? a) \(R / 16\) b) \(R / 8\) c) \(R / 2\) d) \(R\)

Short answer

a) R b) R/4 c) R/2 d) 3R/4 Answer: c) R/2

Step-by-step solution

Write down the known information

The first point is at a distance of \(R / 4\) from the axis of rotation. We need to find the distance of the second point such that its linear velocity is twice that of the first point.

Recall the formula connecting linear and angular velocities

The formula that connects linear velocity (v) and angular velocity (ω) is: v = rω where r is the distance from the axis of rotation.

Write down the velocities of both points

Let \(v_1\) be the linear velocity of the first point and \(v_2\) be the linear velocity of the second point. Since the second point has a linear velocity twice that of the first point, we have: \(v_2 = 2v_1\)

Write down the distances from the axis of rotation

Let \(r_1\) be the distance of the first point from the axis of rotation, and \(r_2\) be the distance of the second point from the axis of rotation. We know that: \(r_1 = R / 4\) We need to find \(r_2\).

Use the formula to find the ratio ω₁/ω₂

We know that \(v_1 = r_1ω_1\) and \(v_2 = r_2ω_2\). Since \(v_2 = 2v_1\), we can write: \(r_2ω_2 = 2r_1ω_1\) Now, since both points are on the same disc, they have the same angular velocity, so \(ω_1 = ω_2\). Therefore, we can write: \(r_2 = 2r_1\)

Solve for r₂

Now that we have the equation, we can solve for \(r_2\). We know \(r_1 = R/4\), so we can substitute that into the equation: \(r_2 = 2(R/4) = R/2\)

Choose the answer

The distance from the axis of rotation for the second point is \(R/2\). Therefore, the correct answer is (c) \(R / 2\).

Key concepts

Angular Velocity

Understanding angular velocity is critical in many physics problems, especially when dealing with rotating objects. In essence, angular velocity represents how fast something rotates, or in other words, the rate of change of the angular position of an object over time. It is commonly represented by the Greek letter \( \omega \). The formula that links linear velocity (v), angular velocity (\( \omega \)), and the distance to the axis of rotation (r) can be expressed as
\[ v = r\omega \]
This fundamental equation shows that the linear velocity of a point on a rotating object increases with its distance from the axis of rotation—assuming a constant angular velocity. In the problem from the textbook, noting that the second point moved at twice the linear velocity of the first was the key to understanding the relationship between distances from the axis of rotation and their respective velocities.

Axis of Rotation

At the heart of any rotational motion is the axis of rotation. This imaginary line runs through the center of the rotating object and is the point about which all circular motion occurs. The importance of the axis of rotation lies in its role as the point of reference for measuring distances in problems involving circular motion. All points on a rotating disc, wheel, or any circular object have the same angular velocity about this axis.

In the context of the original problem, the axis of rotation provided the reference from which distances \( r_1 \) and \( r_2 \) were measured. Since angular velocity is constant for all points on a rigid rotating object, understanding how the axis of rotation interacts with the concept of angular velocity is crucial in solving circular motion problems.

Circular Motion

Circular motion refers to the movement of an object along the circumference of a circle. It can be uniform, with constant angular velocity, or non-uniform, with varying angular velocity. In our textbook problem, we assume uniform circular motion, where any point on the disc maintains a constant angular velocity.

An intriguing aspect of circular motion is that it always involves acceleration, even in uniform circular motion, because the direction of the velocity vector changes continuously. This acceleration, called centripetal acceleration, points towards the axis of rotation and keeps the object in its circular path. The interplay of linear velocity, angular velocity, and centripetal acceleration all depends on the distance from the axis of rotation—as highlighted in solving the original problem.