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(a) Derive an equation for the radial acceleration that includes \(v\) and \(\omega,\) but not \(r .\) (b) You are designing a merry-go-round for which a point on the rim will have a radial acceleration of 0.500 \(\mathrm{m} / \mathrm{s}^{2}\) when the tangential velocity of that point has magnitude 2.00 \(\mathrm{m} / \mathrm{s}\) . What angular velocity is required to achieve these values?

Short Answer

Expert verified
The required angular velocity is 0.250 rad/s.

Step by step solution

01

Define Radial Acceleration

Radial acceleration, also known as centripetal acceleration, is given by the formula \( a_r = \frac{v^2}{r} \), where \(v\) is the tangential velocity and \(r\) is the radius. It can also be expressed as \( a_r = r\omega^2 \), where \( \omega \) is the angular velocity.
02

Derive Equation for Radial Acceleration Without r

To remove \(r\), equate the two expressions for radial acceleration: \(\frac{v^2}{r} = r\omega^2\). Solving for \(r\), we get \(r = \frac{v}{\omega}\). Substitute \(r = \frac{v}{\omega}\) into \( a_r = \omega^2 r\) to obtain: \(a_r = \omega^2 \left(\frac{v}{\omega}\right) = v\omega \). Thus, the equation for radial acceleration without \(r\) is \(a_r = v\omega\).
03

Substitute Known Values

We need to find the angular velocity \(\omega\) given \(a_r = 0.500\,\text{m/s}^2\) and \(v = 2.00\,\text{m/s}\). Substitute these values into the equation \(a_r = v\omega\): \[0.500 = 2.00 \cdot \omega\].
04

Solve for Angular Velocity

Rearrange the equation \(0.500 = 2.00 \cdot \omega\) to find \(\omega\): \[ \omega = \frac{0.500}{2.00} = 0.250\,\text{rad/s} \].

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Key Concepts

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

Centripetal Force
Imagine you are spinning a ball on a string around your head. The force that keeps the ball moving in a circle is called the centripetal force. It acts towards the center of the circle and is responsible for the ball's curved path. Whenever an object moves in a circle or along a curved path, centripetal force ensures that it doesn't fly off in a straight line.
This force is not a separate force of nature but results from other fundamental forces like tension, gravity, or friction. For example:
  • A car making a turn experiences centripetal force from the friction between the tires and the road.
  • The gravitational pull of the Earth provides the centripetal force needed to keep the Moon in its orbit.
  • On a merry-go-round, the structure provides the centripetal force that keeps children safely riding in circular paths.
When we talk about radial acceleration in circular motion, we're referring to how quickly the object's velocity is changing direction, which is directly related to this centripetal force. Without it, circular motion wouldn't be possible!
Angular Velocity
When something rotates or spins, it does so at an angular velocity. Angular velocity measures how quickly an object moves through an angle. Think of it like the speed of a spinning wheel or a rotating fan. It's often measured in radians per second \((rad/s)\).
Angular velocity is crucial when describing rotational motion because it tells us how fast something is spinning. For example:
  • When you wash clothes in a washing machine, the drum spins with a certain angular velocity to remove water by centrifugal forces.
  • A spinning top maintains its stability because of its angular motion.
The formula relating angular velocity to tangential (or linear) velocity is:\[ v = r \cdot \omega \]where \(v\) is the tangential velocity, \(r\) is the radius of the path, and \(\omega\) is the angular velocity. Changing the angular velocity changes how rapidly different points on the rotating object move through space. A high angular velocity means faster circular movement, while a low angular velocity means slower motion.
Tangential Velocity
Tangential velocity is the speed at which an object moves along its circular path. Imagine a point on the edge of a spinning merry-go-round. This point moves along the circular path, and its speed is the tangential velocity. It's called 'tangential' because it is always directed along the tangent to the path of the circle.
The key thing to remember about tangential velocity is:
  • It depends on both the radius of the circular path and the angular velocity.
  • Larger radius and higher angular velocity lead to greater tangential velocity.
The relationship between tangential velocity \(v\), radius \(r\), and angular velocity \(\omega\) is:\[ v = r \cdot \omega \]This formula shows how changes in angular velocity or radius affect how quickly the object is moving along the circle. For someone sitting near the edge of a carousel, the ride feels faster because their tangential velocity is much higher than someone closer to the center.
Understanding tangential velocity is crucial for analyzing motions in circular paths and can be seen in various everyday activities, from a spinning CD to the orbits of planets around the sun.

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Most popular questions from this chapter

A meter stick with a mass of 0.160 \(\mathrm{kg}\) is pivoted about one end so it can rotate without friction about a horizontal axis. The meter stick is held in a horizontal position and released. As it swings through the vertical, calculate (a) the change in gravitational potential energy that has occurred; (b) the angular speed of the stick; (c) the linear speed of the end of the stick opposite the axis. (d) Compare the answer in part (c) to the speed of a particle that has fallen \(1.00 \mathrm{m},\) starting from rest.

A wheel changes its angular velocity with a constant angular acceleration while rotating about a fixed axis through its center (a) Show that the change in the magnitude of the radial acceleration during any time interval of a point on the wheel is twice the product of the angular acceleration, the angular displacement, and the perpendicular distance of the point from the axis. (b) The radial acceleration of a point on the wheel that is 0.250 \(\mathrm{m}\) from the axis changes from 25.0 \(\mathrm{m} / \mathrm{s}^{2}\) to 85.0 \(\mathrm{m} / \mathrm{s}^{2}\) as the wheel rotates through 15.0 rad. Calculate the tangential acceleration of this point. (c) Show that the change in the wheel's kinetic energy during any time interval is the product of the moment of inertia about the axis, the angular acceleration, and the angular displacement. (d) During, the 15.0 -rad angular displacement of part (b), the kinetic energy \(y\) y of the wheel increases from 20.0 \(\mathrm{J}\) to 45.0 \(\mathrm{J}\) . What is the moment of inertia of the wheel about the rotation axis?

Electric Drill. According to the shop manual, when drilling a \(12.7-\mathrm{mm}\) -diameter hole in wood, plastic, or aluminum, a drill should have a speed of 1250 rev/min. For a 12.7 -mm-diameter drill bit turning at a constant 1250 rev \(/ \mathrm{mm}\) , find (a) the maximum linear speed of any part of the hit and (b) the maximum radial acceleration of any part of the bit.

A fan blade rotates with angular velocity given by \(\omega_{z}(t)=\) \(\gamma-\beta t^{2},\) where \(\gamma=5.00 \mathrm{rad} / \mathrm{s}\) and \(\beta=0.800 \mathrm{rad} / \mathrm{s}^{3} .\) (a) Calculate the angular acceleration as a function of time. (b) Calculate the instantaneous angular acceleration \(\alpha_{z}\) at \(t=3.00 \mathrm{s}\) and the average angular acceleration \(\alpha_{\mathrm{axz}}\) for the time interval \(t=0\) to \(t=3.00 \mathrm{s}\) . How do these two quantities compare? If they are different, why are they different?

(a) What angle in radians is subtended by an arc 1.50 \(\mathrm{m}\) long on the circumference of a circle of radius 2.50 \(\mathrm{m} ?\) What is this angle in degrees? (b) An are 14.0 \(\mathrm{cm}\) long on the circumference of a circle subtends an angle of \(128^{\circ} .\) What is the radius of the circle? (c) The angle between two radii of a circle with radius 1.50 \(\mathrm{m}\) is 0.700 rad. What length of arc is intercepted on the circumference of the circle by the two radii?

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