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Chapter 10: Problem 13

A 2.00 -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 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
Tensions are 7.50 N (Block A) and 18.18 N (Block B); moment of inertia is 0.016 kg路m虏.

Step by step solution

01

Analyze the given data

We have two blocks: Block A (mass = 2.00 kg) on a frictionless surface and Block B (mass = 3.00 kg) hanging. They move a distance of 1.20 m in 0.800 s. The pulley has a diameter of 0.150 m.
02

Calculate acceleration of the system

Using the formula for acceleration, \(a = \frac{2d}{t^2}\), where \(d = 1.20\,\mathrm{m}\) and \(t = 0.800\,\mathrm{s}\). Substituting the values gives: \[ a = \frac{2 \times 1.20}{(0.800)^2} = 3.75\,\mathrm{m/s^2} \]
03

Apply Newton's Second Law to the blocks

For Block A, the tension \(T_1\) can be found using \(T_1 = m_A \cdot a\), where \(m_A = 2.00\,\mathrm{kg}\). For Block B, \(m_Bg - T_2 = m_B \cdot a\), where \(m_B = 3.00\,\mathrm{kg}\) and \(g = 9.81\,\mathrm{m/s^2}\).
04

Solve for tensions in the cord

Substitute the known values in the above equations:- For Block A: \[ T_1 = 2.00 \times 3.75 = 7.50\,\mathrm{N} \] - For Block B: \[ T_2 = 3.00 \cdot 9.81 - 3.00 \cdot 3.75 = 18.18\,\mathrm{N} \]
05

Evaluate the net torque on the pulley

The torque on the pulley is given by \( \tau = (T_2 - T_1) \times r \), where \(r = \frac{0.150}{2} = 0.075\,\mathrm{m}\). Substituting for \(T_2\) and \(T_1\), we get:\[ \tau = (18.18 - 7.50) \times 0.075 = 0.801\,\mathrm{N\cdot m} \]
06

Solve for the moment of inertia of the pulley

Use the relation \( \tau = I \cdot \alpha \), where \( \alpha = \frac{a}{r} = \frac{3.75}{0.075}\). Continuing from the torque equation:\[ I = \frac{\tau}{\alpha} = \frac{0.801}{50} = 0.016\,\mathrm{kg\cdot m^2} \]

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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 is fundamental in understanding motion. It states that the acceleration of an object is directly proportional to the net force acting on it, and inversely proportional to its mass. The formula is expressed as \( F = m \cdot a \), where \( F \) is the force, \( m \) is the mass, and \( a \) is the acceleration.
In this exercise, the net forces acting on two blocks connected by a cord influence their acceleration. The key is to understand that each block experiences different forces due to their position - one being on a frictionless surface and the other hanging.
This law helps us calculate the tension in the cords and ultimately determines how forces are balanced in the system of pulleys and blocks.
Tension in Cords
Understanding tension is crucial when dealing with pulleys and cords. Tension refers to the force transmitted through a string or rope when it is pulled tight by forces acting from opposite ends.
In this instance, two different tensions, \( T_1 \) and \( T_2 \), occur due to different masses on each side of the pulley.
For Block A, the tension \( T_1 \) is calculated using the formula \( T_1 = m_A \cdot a \), where \( a \) is the acceleration. This represents the force needed to maintain the motion of Block A on the surface.
Conversely, for Block B, tension \( T_2 \) is derived using the equation \( m_Bg - T_2 = m_B \cdot a \), which considers the gravitational force pulling Block B downwards. Calculating these tensions is vital for understanding how forces distribute within mechanical systems.
Moment of Inertia
The moment of inertia is a concept relating to rotational motion, akin to mass in linear motion. It measures an object's resistance to change in its rotational motion.
For a disc-like pulley, it depends on its mass distribution relative to the rotation axis. The given exercise involves determining the pulley's moment of inertia by analyzing the torque created by the tensions \( T_1 \) and \( T_2 \) acting at the pulley's radius.
The formula \( \tau = I \cdot \alpha \) (torque = moment of inertia \( I \) times angular acceleration \( \alpha \)) connects these quantities.
Understanding this helps us appreciate how inertia affects the rotational dynamics of mechanical systems, particularly in devising effective designs for pulleys.
Acceleration Calculation
To determine how quickly the books move, we calculate their acceleration. The formula \( a = \frac{2d}{t^2} \) helps find this, where \( d \) is the distance traveled and \( t \) is the time taken.
From the exercise data, the acceleration is calculated as \[ a = \frac{2 \times 1.20}{(0.800)^2} = 3.75\,\mathrm{m/s^2} \]. This step establishes the rate at which the textbooks' speed changes.
This calculated acceleration is used to solve for the tension in the cords and the rotational characteristics of the pulley.
Understanding acceleration in physics allows students to link how forces acting on the motion of objects result in certain behaviors, bridging the gap between theoretical principles and real-world applications.

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Most popular questions from this chapter

A hollow, spherical shell with mass 2.00 \(\mathrm{kg}\) rolls without slipping down a \(38.0^{\circ}\) slope. (a) Find the acceleration, the friction force, and the minimum coefficient of friction needed to prevent slipping. (b) How would your answers to part (a) change if the mass were doubled to 4.00 \(\mathrm{kg}\) ?

A uniform, \(0.0300-\mathrm{kg}\) rod of length 0.400 \(\mathrm{m}\) rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.0200 \(\mathrm{kg}\) , are mounted so that they can slide along the rod. They are initially held by catches at positions 0.0500 \(\mathrm{m}\) on each side of the center of the rod, and the system is rotating at 30.0 rev/min. With no other changes in the system, the catches are released, and the rings slide outward along the rod and fly off at the ends. (a) What is the angular speed of the system at the instant when the rings reach the ends of the rod? (b) What is the angular speed of the rod after the rings leave it?

While exploring a castle, Exena the Exterminator is spotted by a dragon that chases her down a hallway. Exena runs into a room and attempts to swing the heavy door shut before the dragon gets her. The door is initially perpendicular to the wall, so it must be turned through \(90^{\circ}\) to close. The door is 3.00 \(\mathrm{m}\) tall and 1.25 \(\mathrm{m}\) wide, and it weighs 750 \(\mathrm{N} .\) You can ignore the friction at the hinges. If Exena applies a force of 220 \(\mathrm{N}\) at the edge of the door and perpendicular to it, how much time does it take her to close the door?

A cord is wrapped around the rim of a solid uniform wheel 0.250 \(\mathrm{m}\) in radius and of mass 9.20 \(\mathrm{kg} .\) A steady horizontal pull of 40.0 \(\mathrm{N}\) to the right is exerted on the cord, pulling it off tangentially from the wheel. The wheel is mounted on frictionless bearings on a horizontal axle through its center. (a) Compute the angular acceleration of the wheel and the acceleration of the part of the cord that has already been pulled off the wheel. (b) Find the magnitude and direction of the force that the axle exerts on the wheel. (c) Which of the answers in parts (a) and (b) would change if the pull were upward instead of horizontal?

A hollow, thin-walled sphere of mass 12.0 \(\mathrm{kg}\) and diameter 48.0 \(\mathrm{cm}\) is rotating about an axle through its center. The angle (in radians) through which it turns as a function of time (in seconds) is given by \(\theta(t)=A t^{2}+B t^{4},\) where \(A\) has numerical value 1.50 and \(B\) has numerical value 1.10 . (a) What are the units of the constants \(A\) and \(B ?\) (b) At the time 3.00 s, find (i) the angular momentum of the sphere and (ii) the net torque on the sphere.
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