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A \(2.00-\mathrm{kg}\) textbook rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is \(0.150 \mathrm{m},\) to a hanging book with mass 3.00 \(\mathrm{kg}\) . The system is released from rest, and the books are observed to move 1.20 \(\mathrm{m}\) in 0.800 \(\mathrm{s}\) s. (a) What is the tension in each part of the cord? \((b)\) What is the moment of inertia of the pulley about its rotation axis?

Short Answer

Expert verified
(a) Tensions are 18.18 N and 7.50 N. (b) Moment of inertia is 0.016 kg m².

Step by step solution

01

Determine the Acceleration

Calculate the acceleration using the kinematic equation: \[ s = ut + \frac{1}{2}at^2 \]where:- \( s = 1.20 \, \text{m} \) is the distance moved,- \( u = 0 \) because the initial speed is zero,- \( t = 0.800 \, \text{s} \) is the time.Solving for \( a \):\[ 1.20 = 0 + \frac{1}{2}a(0.800)^2 \]\[ a = \frac{2 \times 1.20}{(0.800)^2} \approx 3.75 \, \text{m/s}^2 \]
02

Apply Newton's Second Law to the Hanging Book

Use Newton's second law \( F = ma \) for the hanging book:The forces are:- Tension \( T \) acting upwards,- Weight \( m_2g \) acting downwards where \( m_2 = 3.00 \mathrm{\, kg} \).Thus, \[ m_2g - T = m_2a \]\[ T = m_2g - m_2a \]Substitute the known values:\[ T = 3.00 \times 9.81 - 3.00 \times 3.75 \]\[ T = 18.18 \, \text{N} \]
03

Apply Newton's Second Law to the Book on the Table

For the book on the frictionless table, tension is the only force:\[ T = m_1a \]where \( m_1 = 2.00 \mathrm{\, kg} \).Substitute \( a = 3.75 \, \text{m/s}^2 \) to find the tension \( T_1 \):\[ T_1 = 2.00 \times 3.75 = 7.50 \, \text{N} \]
04

Determine Pulley's Moment of Inertia

To find the pulley's moment of inertia, consider the net torque on the pulley:Net torque is caused by the difference in tensions \( T_2 - T_1 \) with radius \( r = \frac{d}{2} = 0.075 \, \text{m} \):\[ (T_2 - T_1)r = I \alpha \]The angular acceleration \( \alpha = \frac{a}{r} \). Hence,\[ (18.18 - 7.50)\times 0.075 = I \times \frac{3.75}{0.075} \]Solving for \( I \):\[ I = \frac{(18.18 - 7.50)\times 0.075}{\frac{3.75}{0.075}} \]\[ I = \frac{(10.68 \times 0.075)}{50} \approx 0.016 \, \text{kg m}^2 \]

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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, defined by the equation \( F = ma \), is a fundamental principle in physics that describes how force impacts motion. It states that the force \( F \) applied on an object is equal to the objects mass \( m \) multiplied by the acceleration \( a \) it experiences.
This law explains the relationship between an object's mass, the force needed to move it, and the resulting acceleration. In the context of our problem, the hanging book, experiencing both gravitational and tension forces, helps illustrate this concept.
  • The gravitational force acts downward, calculated as the product of mass \( m_2 \) and gravity \( g \), i.e., \( m_2g \).
  • Tension force acts upward and counters part of the gravitational pull.
By applying the principle \( m_2g - T = m_2a \), we can determine the tension in the cord affecting the hanging mass. This illustrates not only the utility of Newton's second law but also its integral role in solving dynamics problems where multiple forces work together.
moment of inertia
Moment of inertia is a measure of an object's resistance to changes in its rotational motion. It plays a role analogous to mass in linear motion. A greater moment of inertia means more torque is required to change the rotational speed of the object.
In the given exercise, the focus is on finding the moment of inertia \( I \) of the pulley, which is influenced by the tension difference in the cord that causes the pulley to rotate.
  • The net torque acting on the pulley is the difference in tensions \( (T_2 - T_1) \) multiplied by the pulley's radius \( r \).
  • Angular acceleration \( \alpha \) is related to linear acceleration \( a \) through the radius, where \( \alpha = \frac{a}{r} \).
Using the relation \( I \alpha = (T_2 - T_1)r \), we solve for \( I \) and understand that \( I = \frac{(T_2 - T_1)r}{\alpha} \), quantifying the pulley's capability to resist rotational acceleration. Moment of inertia thus becomes critical for predicting how rotational motion evolves under force.
kinematic equations
Kinematic equations help describe the motion of objects and are essential in finding unknown variables like distance, velocity, acceleration, and time. The equation \( s = ut + \frac{1}{2}at^2 \) was used in the exercise to determine the acceleration \( a \) of the system as it started from rest.
Here are the elements involved in this kinematic equation:
  • \( s \) is the displacement, which is given as 1.20 m.
  • Initial velocity \( u \) is zero since the system starts from rest.
  • \( t \) denotes the time interval, which is 0.800 seconds.
Solving this equation gives us the required acceleration needed in applying Newton's law. It implies that by knowing the initial conditions and displacement over time, one can predict the kinematic outcomes in dynamic systems. Kinematics thus provides the mathematical tools to bridge the initial conditions with dynamic laws like Newton's, enabling comprehensive analysis even in more complex scenarios.
rotational dynamics
Rotational dynamics deals with the rotational motion of objects, focusing on the effects of torques and moments of inertia. When objects rotate, every force and the mass distribution both play a role in determining how they spin.
In our problem, the motion of the pulley is dictated by:
  • The difference in tension on either side of the pulley applying a torque.
  • The moment of inertia, which defines the pulley's rotational inertia.
  • The rotational analog of Newton's second law, \( I \alpha = \tau \), where \( \tau \) is torque.
Understanding rotational dynamics requires insights from both torque interactions and the moment of inertia. The angular acceleration \( \alpha \) and how it is derived from linear motion \( \alpha = \frac{a}{r} \) forms part of this framework. This area of physics can predict how rotations evolve and balance forces in systems with rotating components, offering a complete picture that linear dynamics alone cannot provide.

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Most popular questions from this chapter

In a physics laboratory you do the following ballistic pen- dulum experiment: You shoot a ball of mass \(m\) horizontally from a spring gun with a speed \(v\) . The ball is immediately caught a distance \(r\) below a frictionless pivot by a pivoted catcher assembly of mass \(M\) . The moment of inertia of this assembly about its rotation axis through the pivot is \(L\) . The distance \(r\) is much greater than the radius of the ball. (a) Use conservation of angular momentum to show that the angular speed of the ball and catcher just after the ball is caught is \(\omega=\operatorname{mor} /\left(m r^{2}+I\right) .\) (b) After the ball is caught, the center of mass of the ball-catcher assembly system swings up with a maximum height increase \(h\) . Use conservation of energy to show that \(\omega=\sqrt{2(M+m) g h /\left(m r^{2}+1\right)} .\) (c) Your lab partner says that linear momentum is conserved in the collision and derives the expression \(m v=(m+M) V,\) where \(V\) is the speed of the ball immediately after the collision. She then uses conservation of energy to derive that \(V=\sqrt{2 g h},\) so that \(m v=(m+M) \sqrt{2 g h} .\) Use the results of parts (a) and (b) to show that this equation is satisfied only for the special case when \(r\) is given by \(I=M r^{2} .\)

A stiff uniform wire of mass \(M_{0}\) and length \(L_{0}\) is cut, bent, and the parts soldered together so that it forms a circular wheel having four identical spokes coming out from the center. None of the wire is wasted, and you can neglect the mass of the solder. (a) What is the moment of inertia of this wheel about an axle through its center perpendicular to the plane of the wheel? (b) If the wheel is given an initial spin with angular velocity \(\omega_{0}\) and stops uniformly in time \(T\) , what is the frictional torque at its axle?

Neutron Star Glitches. Occasionally, a rotating neutron star (see Exercise 10.39 ) undergoes a sudden and unexpected speedup called a glitch. One explanation is that a ghitch occurs when the crust of the neutron star settles slightly, decreasing the moment of inertia about the rotation axis. A neutron star with angular speed \(\omega_{0}=70.4\) rad/s underwent such a glitch in October 1975 that increased its angular speed to \(\omega=\omega_{0}+\Delta \omega,\) where \(\Delta \omega / \omega_{0}=2.01 \times 10^{-6} .\) If the radius of the neutron star before the glitch was \(11 \mathrm{km},\) by how much did its radius decrease in the star-quake? Assume that the neutron star is a uniform sphere.

A stone is suspended from the free end of a wire that is wrapped around the outer rim of a pulley, similar to what is shown in Fig. \(10.10 .\) The pulley is a uniform disk with mass 10.0 \(\mathrm{kg}\) and radius 50.0 \(\mathrm{cm}\) and turns on frictionless bearings. You measure that the stone travels 12.6 \(\mathrm{m}\) in the first 3.00 s starting from rest. Find (a) the mass of the stone and (b) the tension in the wire.

The flywheel of an engine has moment of inertia 2.50 \(\mathrm{kg} \cdot \mathrm{m}^{2}\) about its rotation axis. What constant torque is required to bring it up to an angular speed of 400 \(\mathrm{rev} / \mathrm{min}\) in 8.00 \(\mathrm{s}\) s, starting from rest?

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