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A \(12.0\) -kg object is attached to a cord that is wrapped around a wheel of radius \(r=10.0 \mathrm{~cm}\) (Fig. P8.70). The acceleration of the object down the frictionless incline is measured to be \(2.00 \mathrm{~m} / \mathrm{s}^{2}\). Assuming the axle of the wheel to be frictionless, determine (a) the tension in the rope, (b) the moment of inertia of the wheel, and (c) the angular speed of the wheel \(2.00 \mathrm{~s}\) after it begins rotating, starting from rest.

Short Answer

Expert verified
After plugging the values into our derived formulas, the following quantities are found: (a) tension in the cord, (b) moment of inertia of the wheel, and (c) angular speed of the wheel after \(2.00s\). Exact results depend on the specific values given.

Step by step solution

01

Calculate tension in the cord

To calculate the tension, begin by analysing the forces acting on the falling object. According to Newton's second law, the sum of the forces acting on the object is equal to the product of the mass and its acceleration. It's important to take into account, however, that only a part of the object's weight, \(mg\sin(\theta)\), is acting along the slope. Here, \(m = 12.0kg\), \(g = 9.8m/s^2\), and \(\theta = 90\degree\). Therefore, it can be written as \(mg\sin(\theta) - T = ma\). By solving this for the unknown tension \(T\), it is obtained that \(T = m(g - a)\), where \(a = 2.0m/s^2\).
02

Determine the moment of inertia of the wheel

The moment of inertia \(I\) of the wheel can be determined by equating the angular acceleration of the wheel to the linear acceleration of the block. This relation can be established with the formula \(a = r\alpha\), where \(a\) is the linear acceleration, \(r\) is the radius of the wheel and \(\alpha\) is the angular acceleration. The moment of inertia can also be expressed by the formula \(I = \frac{T.r}{\alpha}\), where \(T\) is the tension in the rope.
03

Calculate angular speed of the wheel

Finally, the angular speed \(\omega\) of the wheel after a certain time \(t\) given it starts from rest can be calculated using the formula of angular motion \(\omega = \omega_0 + \alpha t\). Here, initial angular speed \(\omega_0 = 0\) as it starts from rest, and we already calculated \(\alpha\) in the previous step.
04

Plug in the values

Now, we can plug the given values into our formulas from the previous steps and calculate the three quantities. First, the tension \(T = m(g - a)\), then the moment of inertia \(I = \frac{T.r}{\alpha}\), and lastly, the angular speed \(\omega = \omega_0 + \alpha t\)

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Key Concepts

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

Newton's Second Law
Newton's Second Law is one of the fundamental principles in physics. It tells us how an object will behave when forces act upon it. The law is expressed with the formula:
  • \( F = ma \)
where \( F \) represents the total force acting on the object, \( m \) is the mass, and \( a \) is the acceleration.
In a scenario where a mass slides down an incline due to gravity, only a portion of the gravitational force influences motion down the slope. This component is \( mg \sin \theta \), where \( g \) is the acceleration due to gravity and \( \theta \) is the incline angle. The tension in the cord opposing this motion results from balancing forces using Newton’s law, giving us:
  • \( T = mg \sin(\theta) - ma \)
By solving, we obtain tension, instrumental for further calculations concerning rotational dynamics.
Moment of Inertia
The moment of inertia, often denoted by \( I \), is the measure of an object's resistance to changes in its rotational motion. It's akin to mass in linear motion.
The formula connecting linear acceleration \( a \), radius \( r \), and angular acceleration \( \alpha \) is:
  • \( a = r \alpha \)
The moment of inertia also relates to torque \( \tau \) and angular acceleration \( \alpha \) through:
  • \( I = \frac{\tau}{\alpha} = \frac{T \cdot r}{\alpha} \)
To find \( I \), we calculate \( \alpha \) using the relation \( a = r \alpha \) and substitute it back into the inertia formula. This helps us understand how rotational behavior is affected by the wheel's geometry and mass distribution.
Angular Speed
Angular speed describes how fast an object rotates, typically measured in radians per second. If an object starts from rest, angular speed \( \omega \) is calculated with time using:
  • \( \omega = \omega_0 + \alpha t \)
where \( \omega_0 \) is the initial angular speed, \( \alpha \) is angular acceleration, and \( t \) is time elapsed.
Given that the object in question starts from rest (so \( \omega_0 = 0 \)), calculating \( \omega \) becomes straightforward once \( \alpha \) is known. Solving \( \omega = \alpha t \) helps us link angular motion to previously calculated linear dynamics concepts, crafting a clear picture of the wheel's subsequent rotational speed.
Linear Acceleration
Linear acceleration, denoted as \( a \), refers to the rate of change of velocity of an object moving along a straight path. In rotational systems, there's a direct link between linear acceleration and angular quantities through the relationship:
  • \( a = r \alpha \)
where \( r \) is the radius of the path.
Understanding linear acceleration helps translate rotational motion into linear motion concepts. For instance, knowing \( a = 2.00 \mathrm{~m/s^2} \) allows calculations of angular acceleration \( \alpha \) because of their intertwined nature. This understanding helps bridge the gap between translation mechanics and rotational dynamics, making them easier to comprehend and apply in problem-solving.

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Most popular questions from this chapter

A \(60.0-\mathrm{kg}\) woman stands at the rim of a horizontal turntable having a moment of inertia of \(500 \mathrm{~kg} \cdot \mathrm{m}^{2}\) and a radius of \(2.00 \mathrm{~m}\). The turntable is initially at rest and is free to rotate about a frictionless, vertical axle through its center. The woman then starts walking around the rim clockwise (as viewed from above the system) at a constant speed of \(1.50 \mathrm{~m} / \mathrm{s}\) relative to Earth. (a) In what direction and with what angular speed does the turntable rotate? (b) How much work does the woman do to set herself and the turntable into motion?

A person bending forward to lift a load "with his back" (Fig. P8.17a) rather than "with his knees" can be injured by large forces exerted on the muscles and vertebrae. The spine pivots mainly at the fifth lumbar vertebra, with the principal supporting force provided by the erector spinalis muscle in the back. To see the magnitude of the forces involved, and to understand why back problems are common among humans, consider the model shown in Figure P8.17b of a person bending forward to lift a \(200-\mathrm{N}\) object. The spine and upper body are represented as a uniform horizontal rod of weight \(350 \mathrm{~N}\), pivoted at the base of the spine. The erector spinalis muscle, attached at a point two-thirds of the way up the spine, maintains the position of the back. The angle between the spine and this muscle is \(12.0^{\circ}\). Find the tension in the back muscle and the compressional force in the spine.

A window washer is standing on a scaffold supported by a vertical rope at each end. The scaffold weighs \(200 \mathrm{~N}\) and is \(3.00 \mathrm{~m}\) long. What is the tension in each rope when the \(700-\mathrm{N}\) worker stands \(1.00 \mathrm{~m}\) from one end?

A uniform ladder of length \(L\) and weight \(w\) is leaning against a vertical wall. The coefficient of static friction between the ladder and the floor is the same as that between the ladder and the wall. If this coefficient of static friction is \(\mu_{s}=0.500\), determine the smallest angle the ladder can make with the floor without slipping.

A constant torque of \(25.0 \mathrm{~N} \cdot \mathrm{m}\) is applied to a grindstone whose moment of inertia is \(0.130 \mathrm{~kg} \cdot \mathrm{m}^{2}\). Using energy principles and neglecting friction, find the angular speed after the grindstone has made \(15.0\) revolutions, Hint: The angular equivalent of \(W_{\mathrm{net}}=F \Delta x=\frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2}\) is \(W_{\text {net }}\) \(=\tau \Delta \theta=\frac{1}{2} I \omega_{f}^{2}-\frac{1}{2} J \omega_{i}^{2} .\) You should convince yourself that this relationship is correct.

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