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(I) What is the angular momentum of a 0.270-kg ball revolving on the end of a thin string in a circle of radius 1.35 m at an angular speed of 10.4 rad/s?

Short Answer

Expert verified
The angular momentum is approximately 5.11 kg⋅m²/s.

Step by step solution

01

Understanding Angular Momentum

Angular momentum for a particle moving in a circular path is calculated using the formula: \( L = I imes ext{angular speed} \), where \( I \) is the moment of inertia for a point mass, which is \( I = m imes r^2 \). Here, \( m \) is the mass of the ball, and \( r \) is the radius of the circle.
02

Calculate the Moment of Inertia

Substitute the given values into the moment of inertia formula:\( I = m imes r^2 = 0.270 imes (1.35)^2 \).Calculate:\[ I = 0.270 imes 1.8225 = 0.491175 ext{ kg} imes ext{m}^2 \]
03

Calculate Angular Momentum

Substitute the computed moment of inertia and the given angular speed into the angular momentum formula:\( L = I imes ext{angular speed} = 0.491175 imes 10.4 \).Calculate:\[ L = 5.10702 ext{ kg} imes ext{m}^2/ ext{s} \]
04

Final Result

The angular momentum of the ball is approximately \( 5.11 ext{ kg} imes ext{m}^2/ ext{s} \) when rounded to three significant figures.

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Key Concepts

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

Moment of Inertia
The moment of inertia is a key concept in understanding how objects behave in rotational motion. It is essentially the rotational equivalent of mass in linear motion. More mass "farther" from the axis means greater inertia.
- **Definition**: The moment of inertia (\(I\)) measures an object's resistance to changes in its rotational motion about an axis.- **Formula for Point Mass**: For a point mass, the formula is\(I = m \times r^2\), where\(m\) is the mass and\(r\) is the distance from the axis of rotation. In the exercise, we have a 0.270-kg ball rotating at a radius of 1.35 m. By substituting these into the formula, we calculate its moment of inertia as 0.491 kg·m².
This means the ball "resists" rotational change, analogous to how a heavier object resists being pushed. Understanding this idea helps appreciate why different objects spin differently depending on their mass distribution.
Angular Speed
Angular speed refers to how quickly an object rotates or revolves relative to another point, usually the center of a circle.
- **Units and Definition**: It is measured in radians per second (rad/s). - **Relation to Circular Motion**: It describes how fast an angle changes over time as an object travels around a path. In our example, the ball rotates at an angular speed of 10.4 rad/s. Angular speed is critical in deciding how fast something is spinning without focusing on a changing radius.
To visualize, imagine a merry-go-round. The seats on the outer edge have the same angular speed as those nearer the center, even though they may travel a greater distance in the same amount of time.
Understanding angular speed gives insight into movement dynamics without needing to track distance or position, just speed in rotation.
Circular Motion
Circular motion encompasses any movement where an object travels in a circular path at a consistent distance from a central point.
- **Centripetal Force**: Keeps objects moving in circular paths by acting towards the center. - **Occurs in two states**: Uniform, where the speed is constant, and non-uniform, where the speed varies. In circular motion, the object’s velocity is always changing, even if its speed remains constant, due to the change in direction.
Using the scenario of the ball spinning in a circular path, it's important to understand not just speed but also the force that acts to keep it on that path. This force ensures the ball doesn’t fly off in a straight line and continues its circular trajectory.
Studying circular motion is crucial because it underpins many physical phenomena from planetary orbits to the functioning of engines and fans.

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Most popular questions from this chapter

(II) A 0.72-m-diameter solid sphere can be rotated about an axis through its center by a torque of 10.8 m\(\cdot\)N which accelerates it uniformly from rest through a total of 160 revolutions in 15.0 s. What is the mass of the sphere?

(III) A hammer thrower accelerates the hammer \((mass = 7.30 kg)\) from rest within four full turns (revolutions) and releases it at a speed of 26.5 m/s. Assuming a uniform rate of increase in angular velocity and a horizontal circular path of radius 1.20 m, calculate \((a)\) the angular acceleration, \((b)\) the (linear) tangential acceleration, \((c)\) the centripetal acceleration just before release, \((d)\) the net force being exerted on the hammer by the athlete just before release, and \((e)\) the angle of this force with respect to the radius of the circular motion. Ignore gravity.

Water drives a waterwheel (or turbine) of radius \(R = 3.0 m\) as shown in Fig. 8-66. The water enters at a speed \(\upsilon_1 = 7.0 m/s\) and exits from the waterwheel at a speed \(\upsilon_2 = 3.8 m/s\). \((a)\) If 85 kg of water passes through per second, what is the rate at which the water delivers angular momentum to the waterwheel? \((b)\) What is the torque the water applies to the waterwheel? \((c)\) If the water causes the waterwheel to make one revolution every 5.5 s, how much power is delivered to the wheel?

A merry-go-round with a moment of inertia equal to \(1260 kg\cdot m^2\) and a radius of 2.5 m rotates with negligible friction at 1.70 rad/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 1.35 rad/s. What is her mass?

(II) A centrifuge rotor rotating at 9200 rpm is shut off and is eventually brought uniformly to rest by a frictional torque of 1.20 m\(\cdot\)N. If the mass of the rotor is 3.10 kg and it can be approximated as a solid cylinder of radius 0.0710 m, through how many revolutions will the rotor turn before coming to rest, and how long will it take?

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