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Chapter 11: Problem 8

( \(a\) ) What is the angular momentum of a figure skater spinning at \(2.8 \mathrm{rev} / \mathrm{s}\) with arms in close to her body, assuming her to be a uniform cylinder with a height of \(1.5 \mathrm{~m},\) a radius of \(15 \mathrm{~cm},\) and a mass of \(48 \mathrm{~kg} ?(b)\) How much torque is required to slow her to a stop in \(5.0 \mathrm{~s}\), assuming she does not move her arms?

Short Answer

Expert verified
(a) 9.52 kg路m虏/s, (b) -5.97 Nm

Step by step solution

01

Convert revolutions per second to radians per second

The figure skater is spinning at a speed of \(2.8\) revolutions per second. To convert this to radians per second, we use the conversion \(1 \text{ revolution} = 2\pi \text{ radians}\). Thus, her angular velocity \(\omega = 2.8 \times 2\pi = 5.6\pi \text{ rad/s}\).
02

Calculate the moment of inertia

The skater is assumed to be a uniform cylinder, so we use the formula for the moment of inertia of a cylinder: \[ I = \frac{1}{2} m r^2 \] where \(m = 48 \text{ kg}\) is the mass, and \(r = 0.15 \text{ m}\) is the radius. Therefore, \[ I = \frac{1}{2} \times 48 \times (0.15)^2 = 0.54 \text{ kg} \cdot \text{m}^2 \].
03

Calculate angular momentum

Angular momentum \(L\) is given by the equation \( L = I \omega \). Using the moment of inertia from Step 2 and the angular velocity from Step 1, \[ L = 0.54 \times 5.6\pi \approx 9.52 \text{ kg} \cdot \text{m}^2/\text{s} \].
04

Find the change in angular velocity needed to stop

The skater needs to go from her initial angular velocity \(5.6\pi\) rad/s to \(0\) rad/s to stop. Therefore, the change in angular velocity \(\Delta\omega = -5.6\pi\) rad/s.
05

Use angular kinematics to find required torque

Torque \(\tau\) can be found using the relation \(\tau = I \alpha\), where \(\alpha\) is the angular acceleration. First, we find \(\alpha\) using the formula \(\alpha = \frac{\Delta\omega}{\Delta t}\), where \(\Delta t = 5.0\) s is the time to stop. \[ \alpha = \frac{-5.6\pi}{5.0} \approx -3.52\pi \text{ rad/s}^2 \] Then calculate the torque: \[ \tau = 0.54 \times -3.52\pi \approx -5.97 \text{ Nm} \].

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Key Concepts

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

Moment of Inertia
Moment of inertia is like the rotational equivalent of mass in linear motion. It tells us how much resistance an object offers to changes in its rotational motion.
The formula typically used for a uniform cylinder is:
  • \( I = \frac{1}{2} m r^2 \)
where \( I \) is the moment of inertia, \( m \) is the mass, and \( r \) is the radius of the cylinder.
This equation highlights that a larger mass or radius will increase the moment of inertia, making it harder to change the object's rotational state. In our exercise, the skater's arms being close to her body reduces the radius, thus minimizing her moment of inertia, which in turn allows her to spin faster.
Angular Velocity
Angular velocity measures how fast an object spins or rotates. It represents the angle turned in a given time period.
We usually denote angular velocity by the symbol \(\omega\). The units are often in radians per second. To find it, we need to convert revolutions per second into radians per second using the conversion:
  • \(1 \text{ revolution} = 2\pi \text{ radians}\)
So, a skater spinning at \(2.8\) revolutions per second has an angular velocity of \(5.6\pi\) radians per second. This conversion is crucial in calculating further properties like angular momentum.
Torque
Just like twisting a bottle cap, torque measures how effectively a force causes an object to rotate. Think of it as a rotational force.
Torque is denoted by \(\tau\) and is related to angular acceleration and moment of inertia through the formula:
  • \( \tau = I \alpha \)
Where \( \alpha \) is the angular acceleration. Torque is essentially needed when you want to change the angular velocity of an object, like slowing down a skater. In our example, the value we calculated was \(-5.97\) Nm, indicating that the skater needs to experience this opposing rotational force to come to a stop in 5 seconds.
Angular Acceleration
Angular acceleration is the rate at which an object's angular velocity changes. It's similar to how linear acceleration deals with changes in speed.
We derive angular acceleration, \(\alpha\), using the difference in angular velocities over a certain time interval:
  • \( \alpha = \frac{\Delta\omega}{\Delta t} \)
In this exercise, \(\Delta t\) is the time taken to stop, which is 5 seconds. The skater's angular velocity drops from \(5.6\pi\) to 0, leading to an angular acceleration of approximately \(-3.52\pi\) rad/s虏. The negative sign indicates a deceleration or reduction in spin speed. Understanding this concept is essential for comprehending rotational motion changes.

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Most popular questions from this chapter

(III) Show that the velocity \(\vec{\mathbf{v}}\) of any point in an object rotating with angular velocity \(\overline{\omega}\) about a fixed axis can be written \(\vec{\mathbf{v}}=\vec{\boldsymbol{\omega}} \times \vec{\mathbf{r}}\) where \(\overline{\mathbf{r}}\) is the position vector of the point relative to an origin \(\mathrm{O}\) located on the axis of rotation. Can \(\mathrm{O}\) be anywhere on the rotation axis? Will \(\vec{\mathbf{v}}=\overline{\boldsymbol{\omega}} \times \vec{\mathbf{r}}\) if \(\mathrm{O}\) is located at a point not on the axis of rotation?

A 220-g top spinning at 15 rev/s makes an angle of \(25^{\circ}\) to the vertical and precesses at a rate of 1.00 rev per \(6.5 \mathrm{~s}\). If its \(\mathrm{CM}\) is \(3.5 \mathrm{~cm}\) from its tip along its symmetry axis, what is the moment of inertia of the top?

(II) For what directions of velocity would the Coriolis effect on an object moving at the Earth's equator be zero?

(II) \(\mathrm{A} 220\) -g top spinning at 15 \(\mathrm{rev} / \mathrm{s}\) makes an angle of \(25^{\circ}\) to the vertical and precesses at a rate of 1.00 rev per 6.5 \(\mathrm{s}\) . If its \(\mathrm{CM}\) is 3.5 \(\mathrm{cm}\) from its tip along its symmetry axis, what is the moment of inertia of the top?

Suppose our Sun eventually collapses into a white dwarf, losing about half its mass in the process, and winding up with a radius \(1.0 \%\) of its existing radius. Assuming the lost mass carries away no angular momentum, what would the Sun's new rotation rate be? (Take the Sun's current period to be about 30 days.) What would be its final kinetic energy in terms of its initial kinetic energy of today?
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