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Chapter 11: Problem 1

(1) What is the angular momentum of a \(0.210-\) bg ball rotating on the cnd of a thin string in a circle of radius 1.35 \(\mathrm{m}\) at an angular speed of 10.4 \(\mathrm{rad} / \mathrm{s}\) ?

Short Answer

Expert verified
The angular momentum of the ball is approximately 3.98 kg路m虏/s.

Step by step solution

01

Identify Given Values

Read the problem carefully and note down all the given values:- Mass of the ball, \(m = 0.210\, \text{kg}\).- Radius of the circle, \(r = 1.35\, \text{m}\).- Angular speed, \(\omega = 10.4\, \text{rad/s}\).
02

Recall the Formula for Angular Momentum

The formula for the angular momentum \(L\) of an object moving in a circle is given by:\[L = I \times \omega\]where \(I\) is the moment of inertia and \(\omega\) is the angular speed. For a point mass at a distance \(r\) from the axis of rotation, the moment of inertia is \(I = m \times r^2\).
03

Calculate the Moment of Inertia

Use the formula for a point mass to find the moment of inertia:\[I = m \times r^2\]Substitute the given values:\[I = 0.210\, \text{kg} \times (1.35\, \text{m})^2 = 0.210 \times 1.8225 = 0.382725\, \text{kg} \cdot \text{m}^2\].
04

Calculate Angular Momentum

Now substitute the value of \(I\) and \(\omega\) into the formula for angular momentum:\[L = I \times \omega\]\[L = 0.382725\, \text{kg} \cdot \text{m}^2 \times 10.4\, \text{rad/s} = 3.97834\, \text{kg} \cdot \text{m}^2 \cdot \text{rad/s}\].

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Key Concepts

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

Moment of Inertia
The concept of the moment of inertia is pivotal in understanding how an object's mass distribution affects its rotation. Think of it as the rotational equivalent of mass in linear motion. It determines how much torque is needed for a desired angular acceleration around an axis.
  • A larger moment of inertia means more effort is needed to change the object's rotational motion.
  • For a point mass at a distance from an axis, the moment of inertia is calculated as: \( I = m \times r^2 \).
  • In our example, we calculate the moment of inertia for a 0.210 kg ball rotating in a circle with radius 1.35 m.
  • This yields \( I = 0.210 \, \text{kg} \times (1.35 \, \text{m})^2 = 0.382725 \, \text{kg} \cdot \text{m}^2 \).
This tells us the resistance of the ball to changes in its rotational motion and plays a crucial role in calculating angular momentum.
Angular Speed
Angular speed, often denoted by \( \omega \), is a measure of how fast an object rotates around a given point or axis. It is expressed in radians per second and is crucial in determining the dynamics of circular motion.
  • Angular speed differs from linear speed, which measures how fast an object moves along a path. Instead, angular speed measures the rate of rotation.
  • In our exercise, the ball rotates at an angular speed of 10.4 rad/s, indicating its fast spinning motion.
  • This rotational rate directly influences the calculation of angular momentum, as it is a factor in the formula: \( L = I \times \omega \).
Understanding angular speed helps in analyzing rotational systems, emphasizing how quickly an object's angular position changes over time.
Circular Motion
Circular motion describes the movement of an object along a circular path. This type of motion can be uniform, with a constant angular speed, or non-uniform. Grasping the basics of circular motion is essential for solving problems related to rotation and angular momentum.
  • Key characteristics of circular motion include radius, which defines the size of the circle, and tangential velocity, related to angular speed.
  • For our scenario, the key parameters are the radius of 1.35 meters and the given angular speed.
  • Circular motion links closely with centripetal force, which keeps the object on its circular path.
By connecting these concepts, students can better understand how objects move in circles and how forces impact their rotational behaviors. This holistic understanding empowers them to tackle real-world applications involving rotating systems.

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Most popular questions from this chapter

(II) Two identical particles have equal but opposite momenta, \(\vec{\mathbf{p}}\) and \(-\vec{\mathbf{p}}\) , but they are not traveling along the same line. Show that the total angular momentum of this system does not depend on the choice of origin.

Show that the kinetic energy \(K\) of a particle of mass \(m\), moving in a circular path, is \(K=L^{2} / 2 I,\) where \(L\) is its angular momentum and \(I\) is its moment of inertia about the center of the circle.

A merry-go-round with a moment of inertia equal to \(1260 \mathrm{~kg} \cdot \mathrm{m}^{2}\) and a radius of \(2.5 \mathrm{~m}\) rotates with negligible friction at \(1.70 \mathrm{rad} / \mathrm{s}\). A child initially standing still next to the merry-go-round jumps onto the edge of the platform straight toward the axis of rotation causing the platform to slow to \(1.25 \mathrm{rad} / \mathrm{s}\). What is her mass?

Why might tall narrow SUVs and buses be prone to "rollover"? Consider a vehicle rounding a curve of radius \(R\) on a flat road. When just on the verge of rollover, its tires on the inside of the curve are about to leave the ground, so the friction and normal force on these two tires are zero. The total normal force on the outside tires is \(F_{\mathrm{N}}\) and the total friction force is \(F_{\mathrm{fr}}\). Assume that the vehicle is not skidding. (a) Analysts define a static stability factor SSF \(=w / 2 h\) where a vehicle's "track width" \(w\) is the distance between tires on the same axle, and \(h\) is the height of the CM above the ground. Show that the critical rollover speed is(b) Determine the ratio of highway curve radii (minimum possible) for a typical passenger car with \(\mathrm{SSF}=1.40\) and an SUV with \(\mathrm{SSF}=1.05\) at a speed of \(90 \mathrm{~km} / \mathrm{h}\). $$ v_{\mathrm{C}}=\sqrt{R g\left(\frac{w}{2 h}\right)} $$

(II) A thin rod of mass \(M\) and length \(\ell\) is suspended vertically from a frictionless pivot at its upper end. A mass \(m\) of putty traveling horizontally with a speed \(v\) strikes the rod at its \(\mathrm{CM}\) and sticks there. How high does the bottom of the rod swing?
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