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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) \(R / 16\) c) \(R / 2\) b) \(R / 8\) d) \(R\)

Short Answer

Expert verified
a) \(R/8\) b) \(R/3\) c) \(R/2\) d) \(R\) Answer: c) \(R/2\)

Step by step solution

01

Write down the given information

The first point has a linear velocity \(v_1\) at a distance \(R / 4\) from the axis of rotation. The second point has a linear velocity \(v_2 = 2v_1\). We need to find the distance r for the second point.
02

Write down the formula for linear velocity

The formula for linear velocity is v = rω, where v is linear velocity, r is the distance from the center, and ω is the angular velocity.
03

Set up the equation

We know that \(v_2 = 2v_1\). Using the formula for linear velocity, we have: \((r_2ω) = 2(r_1ω)\) Since both points are on the same Blu-ray disc, the angular velocity ω is the same for both points. That means we can cancel out ω and we're left with: \(r_2 = 2r_1\)
04

Substitute given values

The first point is at a distance of \(R/4\), so \(r_1 = R/4\). Now we can substitute this value into the equation: \(r_2 = 2(R/4)\)
05

Solve for \(r_2\)

Now we simply need to solve for \(r_2\): \(r_2 = R/2\) The distance from the axis of rotation for the second point with twice the linear velocity of the first point is \(\boxed{R/2}\), which is option (c).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Velocity
Understanding linear velocity is crucial when studying motion in physics. It represents the rate at which an object travels along a path, and it is a vector quantity, meaning it has both a magnitude and a direction. The standard unit of linear velocity is meters per second (m/s).

In the context of circular motion, such as with a Blu-ray disc spinning on an axis, each point on the disc can have a different linear velocity depending on its distance from the center of rotation. The formula to find linear velocity is given by the equation:
\( v = r\omega \),
where \( v \) is the linear velocity, \( r \) is the radius or the distance from the axis of rotation to the point of interest, and \( \omega \) is the angular velocity. If we consider the case of two points on a spinning disc, with one point having a linear velocity twice that of the other, we must remember that the angular velocity is constant for all points because they complete a rotation in the same amount of time. However, linear velocity increases with the distance from the axis of rotation. This is why the second point, with twice the linear velocity, must be at a distance from the axis that is twice as far as the first point.
Angular Velocity
Angular velocity, often symbolized as \( \omega \), is the rate of change of the angle through which an object rotates. In simpler terms, it tells us how fast an object spins around its axis of rotation. The unit of angular velocity is radians per second (rad/s). Unlike linear velocity, angular velocity is the same for all points on a rigid rotating body.

The relationship between linear and angular velocity can be understood through an ice skater performing a spin. As the skater pulls her arms in, she reduces the radius of her body's rotation and spins faster. Her angular velocity increases, but because her linear velocity depends on the distance from the axis, it remains constant if she keeps her skates on the same path on the ice. In the case of the Blu-ray disc, both points—the one at \( R/4 \) and the one that we calculate using the linear velocity—share the same angular velocity. This shared angular velocity allows us to create a direct proportion between the linear velocities and the radii of the two points.
Axis of Rotation
The axis of rotation is the straight line around which an object rotates or revolves. Imagine an axle passing through the center of a wheel; this axle serves as the wheel's axis of rotation. In everyday objects, the axis of rotation can be visible, like the hinge on a door, or invisible, such as the geometrical center of a spinning disc.

An object's distance from the axis of rotation is a determining factor for various properties of its motion. For instance, parts of a rotating object farther from the axis of rotation move with higher linear velocities, as highlighted in the exercise with the Blu-ray disc. The axis is pivotal (no pun intended) because it is the reference around which we measure both linear and angular velocities for points on a rotating body. When solving rotational motion problems, identifying the axis of rotation simplifies the relationship between linear velocity, angular velocity, and the object's geometry, giving us a clearer path to solutions such as the one provided with the Blu-ray disc exercise.

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Most popular questions from this chapter

A discus thrower (with arm length of \(1.20 \mathrm{~m}\) ) starts from rest and begins to rotate counterclockwise with an angular acceleration of \(2.50 \mathrm{rad} / \mathrm{s}^{2}\) a) How long does it take the discus thrower's speed to get to \(4.70 \mathrm{rad} / \mathrm{s} ?\) b) How many revolutions does the thrower make to reach the speed of \(4.70 \mathrm{rad} / \mathrm{s} ?\) c) What is the linear speed of the discus at \(4.70 \mathrm{rad} / \mathrm{s}\) ? d) What is the linear acceleration of the discus thrower at this point? e) What is the magnitude of the centripetal acceleration of the discus thrown? f) What is the magnitude of the discus's total acceleration?

A Ferris wheel rotates slowly about a horizontal axis. Passengers sit on seats that remain horizontal on the Ferris wheel as it rotates. Which type of force provides the centripetal acceleration on the passengers when they are at the top of the Ferris wheel? a) centrifugal c) gravity b) normal d) tension

At a county fair, a boy takes his teddy bear on the giant Ferris wheel. Unfortunately, at the top of the ride, he accidentally drops his stuffed buddy. The wheel has a diameter of \(12.0 \mathrm{~m},\) the bottom of the wheel is \(2.00 \mathrm{~m}\) above the ground and its rim is moving at a speed of \(1.00 \mathrm{~m} / \mathrm{s}\). How far from the base of the Ferris wheel will the teddy bear land?

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A flywheel of radius \(31.59 \mathrm{~cm}\) is rotating at a constant frequency. The centripetal acceleration at a point on the edge of the flywheel is \(8.629 \cdot 10^{4} \mathrm{~m} / \mathrm{s}^{2}\). What is the frequency of rotation of the flywheel (in rpm)?

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