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(a) When a disk rolls without slipping, should the product \(r \omega\) be (1) greater than, (2) equal to, or (3) less than \(v_{\mathrm{CM}}\) ? (b) A disk with a radius of \(0.15 \mathrm{~m}\) rotates through \(270^{\circ}\) as it travels \(0.71 \mathrm{~m}\). Does the disk roll without slipping? Prove your answer.

Short Answer

Expert verified
(a) Equal to. (b) Yes, the disk rolls without slipping.

Step by step solution

01

Understanding Rolling Without Slipping

When a disk rolls without slipping, the linear speed of the center of mass, \(v_{\text{CM}}\), is directly related to its angular speed \(\omega\) by the formula \(v_{\text{CM}} = r \omega\). This means that for rolling without slipping, the product \(r \omega\) should be equal to \(v_{\text{CM}}\).
02

Interpreting Given Variables

The problem provides the radius \(r = 0.15\,\mathrm{m}\), the distance traveled \(0.71\,\mathrm{m}\), and the angle of rotation \(\theta = 270^{\circ}\). Convert the angle of rotation from degrees to radians for calculation: \(\theta = \frac{270}{180}\pi = \frac{3\pi}{2}\,\mathrm{radians}\).
03

Calculate the Path of Rotation

Calculate the arc length corresponding to this angle of rotation using the relation \(s = r \theta\), where \(s\) is the arc length. Substituting the given values: \[s = 0.15 \times \frac{3\pi}{2} = 0.707\,\mathrm{m}\].
04

Comparing Distance Traveled and Arc Length

Check if the arc length \(s\) calculated matches the distance traveled by the disk. The calculated arc length \(0.707\,\mathrm{m}\) is very close to the given distance \(0.71\,\mathrm{m}\), suggesting that the disk approximately rolls without slipping since \(s \approx 0.71\,\mathrm{m}\).

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Key Concepts

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

Angular Speed
Angular speed, denoted as \( \omega \), is a measure of how quickly an object rotates. In simple terms, it describes how fast something is spinning around a particular axis. It is measured in radians per second \((\mathrm{rad/s})\).

The formula for angular speed is given by \( \omega = \frac{\Delta \theta}{\Delta t} \), where \( \Delta \theta \) is the change in the rotational angle and \( \Delta t \) is the change in time.

In the context of rolling without slipping, the angular speed of a wheel or disk is directly related to the linear speed at which it moves. Because of this, understanding angular speed is crucial for determining if a body rolls without slipping. When the relationship \( v_{\mathrm{CM}} = r \omega \) holds true, where \( v_{\mathrm{CM}} \) is the linear speed of the center of mass and \( r \) is the radius of the disk, the object rolls without slipping.
Center of Mass
The center of mass of an object is a point that represents the average position of its mass. It is the point where the mass of the body can be thought to be concentrated. This is essential when analyzing movement and rotation.

When an object moves linearly or rotates, its center of mass behaves as if the entire mass were concentrated there. In a rolling motion, especially without slipping, the motion of the center of mass is directly linked to both the linear and angular speeds.

For a disk rolling without slipping, the velocity of the center of mass \( v_{\mathrm{CM}} \) is a key factor and is directly related to the disk's angular speed by \( v_{\mathrm{CM}} = r \omega \). Understanding this relationship helps in determining the conditions needed for the disk to roll correctly without slipping.
Arc Length
Arc length is the distance along the curve of a circle or a circular path. It is an essential concept when working with circles and rotations.

For a given angle \( \theta \) (measured in radians) and radius \( r \), the arc length \( s \) of a circle is given by the formula \( s = r \theta \). This relationship shows how far a point on the circle's edge travels when the circle rotates.

Understanding arc length is crucial in calculating how much a disk or wheel rotates as it moves. In our problem, calculating the arc length ensures we can compare it to the linear distance traveled by the disk. Being able to check if this arc length matches the distance traveled is what allows us to confirm that the disk rolls without slipping. If \( s \approx \text{distance traveled} \), it supports the rolling without slipping condition.
Linear Speed
Linear speed refers to how fast an object moves along a path. It is a measure of the distance traveled per unit of time.

In the scenario of rolling without slipping, linear speed, denoted \( v_{\mathrm{CM}} \) for the center of mass, is an important factor. This type of movement happens when an object rolls such that there is no relative sliding motion between the contact points of the object and the surface.

For a disk rolling without slipping, the linear speed is directly linked to the angular speed through the relationship \( v_{\mathrm{CM}} = r \omega \). This equation ensures that the linear speed at the edge of the rolling body equals the speed of its center of mass. This relationship is fundamental in ensuring the condition of rolling without slipping is achieved, helping to understand the overall system dynamics in this common physics problem.

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Most popular questions from this chapter

Circular disks are used in automobile clutches and transmissions. When a rotating disk couples to a stationary one through frictional force, the energy from the rotating disk can transfer to the stationary one. (a) Is the angular speed of the coupled disks (1) greater than, (2) less than, or (3) the same as the angular speed of the original rotating disk? Why? (b) If a disk rotating at 800 rpm couples to a stationary disk with three times the moment of inertia, what is the angular speed of the combination?

A cylindrical hoop, a cylinder, and a sphere of equal radius and mass are released at the same time from the top of an inclined plane. Using the conservation of mechanical energy, show that the sphere always gets to the bottom of the incline first with the fastest speed and that the hoop always arrives last with the slowest speed.

You wish to accelerate a small merry-go-round from rest to a rotational speed of one-third of a revolution per second by pushing tangentially on it. Assume the merrygo-round is a disk with a mass of \(250 \mathrm{~kg}\) and a radius of \(1.50 \mathrm{~m} .\) Ignoring friction, how hard do you have to push tangentially to accomplish this in \(5.00 \mathrm{~s}\) ? (Use energy methods and assume a constant push on your part.)

A kitten stands on the edge of a lazy Susan (a turntable). Assume that the lazy Susan has frictionless bearings and is initially at rest. (a) If the kitten starts to walk around the edge of the lazy Susan, the lazy Susan will (1) remain lazy and stationary, (2) rotate in the direction opposite that in which the kitten is walking, or (3) rotate in the direction the kitten is walking. Explain. (b) The mass of the kitten is \(0.50 \mathrm{~kg},\) and the lazy Susan has a mass of \(1.5 \mathrm{~kg}\) and a radius of \(0.30 \mathrm{~m}\). If the kitten walks at a speed of \(0.25 \mathrm{~m} / \mathrm{s},\) relative to the ground, what will be the angular speed of the lazy Susan? (c) When the kitten has walked completely around the edge and is back at its starting point, will that point be above the same point on the ground as it was at the start? If not, where is the kitten relative to the starting point? (Speculate on what might happen if everyone on the Earth suddenly started to run eastward. What effect might this have on the length of a day?)

IE A small heavy object of mass \(m\) is attached to a thin string to make a simple pendulum whose length is \(L\) When the object is pulled aside by a horizontal force \(F\) it is in static equilibrium and the string makes a constant angle \(\theta\) from the vertical. (a) The tension in the string should be (1) the same as, (2) greater than, or (3) less than the object's weight, \(m g .\) (b) Use the force condition for static equilibrium (along with a free-body diagram of the object) to prove that the string tension is \(T=\frac{m g}{\cos \theta}>m g\). Use the same procedure to show that \(F=m g \tan \theta\). (c) Prove the same result for \(F\) as in part (b) using the torque condition, summing the torques about the string's tied end. Explain why you cannot use this method to determine the string tension.

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