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Chapter 9: Problem 87

Two metal disks, one with radius \(R_{1}=2.50 \mathrm{cm}\) and mass \(M_{1}=0.80 \mathrm{kg}\) and the other with radius \(R_{2}=5.00 \mathrm{cm}\) and mass \(M_{2}=1.60 \mathrm{kg},\) are welded together and mounted on a frictionless axis through their common center (Fig. P9.87). (a) What is the total moment of inertia of the two disks? (b) A light string is wrapped around the edge of the smaller disk, and a 1.50 -kg block is suspended from the free end of the string. If the block is released from rest at a distance of 2.00 \(\mathrm{m}\) above the floor, what is its speed just before it strikes the floor? (c) Repeat the calculation of part (b), this time with the string wrapped around the edge of the larger disk. In which case is the final speed of the block greater? Explain why this is so.

Short Answer

Expert verified
(a) 2.25 脳 10^{-3} kg路m虏; (b) 5.43 m/s; (c) 6.06 m/s, larger disk results in greater speed.

Step by step solution

01

Calculate Moment of Inertia for Each Disk

The moment of inertia for a disk about its center is given by the formula \( I = \frac{1}{2} m r^2 \). First, calculate the moment of inertia for the smaller disk:\[I_1 = \frac{1}{2} M_1 R_1^2 = \frac{1}{2} \times 0.80 \, \mathrm{kg} \times (0.0250 \, \mathrm{m})^2 = 2.50 \times 10^{-4} \, \mathrm{kg \cdot m^2}\]Next, calculate the moment of inertia for the larger disk:\[I_2 = \frac{1}{2} M_2 R_2^2 = \frac{1}{2} \times 1.60 \, \mathrm{kg} \times (0.0500 \, \mathrm{m})^2 = 2.00 \times 10^{-3} \, \mathrm{kg \cdot m^2}\]
02

Calculate Total Moment of Inertia

The total moment of inertia of the two disks combined about their common center is the sum of their individual moments of inertia:\[I_\text{total} = I_1 + I_2 = 2.50 \times 10^{-4} \, \mathrm{kg \cdot m^2} + 2.00 \times 10^{-3} \, \mathrm{kg \cdot m^2} = 2.25 \times 10^{-3} \, \mathrm{kg \cdot m^2}\]
03

Analyze System with Smaller Disk

Using conservation of energy, start with the gravitational potential energy of the block and end with its kinetic energy and the rotational energy of the disks. Initially, the potential energy is \(U = mgh\), and the final energy is \(\frac{1}{2}mv^2 + \frac{1}{2}I_\text{total}\omega^2\), where \(\omega = \frac{v}{R_1}\) due to the relationship between linear and angular speed.\[mgh = \frac{1}{2} mv^2 + \frac{1}{2} \left(I_\text{total}\frac{v^2}{R_1^2}\right)\]Simplifying, solve for \(v\):\[v = \sqrt{\frac{2mgh}{m + \frac{I_\text{total}}{R_1^2}}}\]Substitute the values:\[v = \sqrt{\frac{2 \times 1.50 \, \mathrm{kg} \times 9.81 \, \mathrm{m/s^2} \times 2.00 \, \mathrm{m}}{1.50 \, \mathrm{kg} + \frac{2.25 \times 10^{-3} \, \mathrm{kg \cdot m^2}}{(0.0250 \, \mathrm{m})^2}}}\]\[v \approx 5.43 \, \mathrm{m/s}\]
04

Analyze System with Larger Disk

Repeat the calculation from Step 3 with \(R_2\) instead of \(R_1\):Using the same approach:\[v = \sqrt{\frac{2mgh}{m + \frac{I_\text{total}}{R_2^2}}}\]Substitute the values:\[v = \sqrt{\frac{2 \times 1.50 \, \mathrm{kg} \times 9.81 \, \mathrm{m/s^2} \times 2.00 \, \mathrm{m}}{1.50 \, \mathrm{kg} + \frac{2.25 \times 10^{-3} \, \mathrm{kg \cdot m^2}}{(0.0500 \, \mathrm{m})^2}}}\]\[v \approx 6.06 \, \mathrm{m/s}\]
05

Compare Final Speeds and Explain

In the case where the block was on the larger disk, the final speed was approximately 6.06 m/s, and on the smaller disk, it was approximately 5.43 m/s. The block achieves a higher speed when wrapped around the larger disk because the moment of inertia affects the energy distribution, and a larger radius results in a smaller fractional contribution of the rotational kinetic energy to total energy, leaving more for translational kinetic energy.

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Key Concepts

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

Conservation of Energy
The principle of conservation of energy is a fundamental concept in physics, implying that energy cannot be created or destroyed, only converted from one form to another. In the context of this exercise, we're looking at how the potential energy of a block transforms into kinetic and rotational energy as it falls. Initially, the block possesses gravitational potential energy due to its height above the ground. As it begins to fall, this energy is converted into two forms:
  • Translational kinetic energy of the moving block.
  • Rotational kinetic energy of the spinning disks attached to the block via the string.
The key here is that the total initial energy (potential energy at the starting height) equals the total final energy (sum of translational and rotational kinetic energies) when the block reaches the ground, assuming no energy losses due to friction or air resistance. This balance of energy allows us to calculate unknown velocities, a powerful tool when analyzing mechanical systems.
Gravitational Potential Energy
Gravitational potential energy (GPE) is the energy held by an object because of its vertical position relative to Earth or another gravitating body. This energy is calculated using the formula:\[ U = mgh \]where:
  • \(m\) is the mass of the object (in kilograms),
  • \(g\) is the acceleration due to gravity (approximately 9.81 m/s虏 on Earth),
  • \(h\) is the height above the reference point (in meters).
In this exercise, the GPE is the energy stored due to the block being suspended at a height. As the block drops, this potential energy is gradually transformed into kinetic energy, propelling the movement of both the block and the disks. The GPE provides the initial energy budget for the entire system to transform and distribute between different forms of energy as the block descends.
Rotational Kinetic Energy
Rotational kinetic energy refers to the portion of the motion energy that is due to an object's rotation. For any rotating object, it is given by the formula:\[ KE_{rot} = \frac{1}{2} I \omega^2 \]where:
  • \(I\) is the moment of inertia, a measure of an object's resistance to changes in its rotation, influenced by mass distribution.
  • \(\omega\) is the angular velocity in radians per second, equivalent to the linear speed divided by the radius of rotation.
In this exercise, two disks are involved, each contributing to the system's moment of inertia. The light string wrapped around one of the disks transmits the falling block's force to the disk, causing it to rotate. When using the smaller disk, the block's energy is more significantly allocated to rotational kinetic energy due to a higher dependency on rotational speed. Conversely, with the larger disk, less energy is used to spin the disk, resulting in a larger portion left for translational kinetic energy, which increases the block's final speed.

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Most popular questions from this chapter

A sphere with radius \(R=0.200 \mathrm{m}\) has density \(\rho\) that decreases with distance \(r\) from the center of the sphere according to \(\rho=3.00 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3}-\left(9.00 \times 10^{3} \mathrm{kg} / \mathrm{m}^{4}\right) r .\) (a) Calculate the total mass of the sphere. (b) Calculate the moment of inertia of the sphere for an axis along a diameter.

At \(t=0\) the current to a dc electric motor is reversed, resulting in an angular displacement of the motor shaft given by \(\theta(t)=(250 \mathrm{rad} / \mathrm{s}) t-\left(20.0 \mathrm{rad} / \mathrm{s}^{2}\right) t^{2}-\left(1.50 \mathrm{rad} / \mathrm{s}^{3}\right) t^{3} .\) (a) \(\mathrm{At}\) what time is the angular velocity of the motor shaft zero? (b) Calculate the angular acceleration at the instant that the motor shaft has zero angular velocity. (c) How many revolutions does the motor shaft turn through between the time when the current is reversed and the instant when the angular velocity is zero? (d) How fast was the motor shaft rotating at \(t=0,\) when the current was reversed? (e) Calculate the average angular velocity for the time period from \(t=0\) to the time calculated in part (a).

A uniform 2.00 -m ladder of mass 9.00 kg is leaning against a vertical wall while making an angle of \(53.0^{\circ}\) with the floor. A worker pushes the ladder up against the wall until it is vertical. What is the increase in the gravitational potential energy of the ladder?

Engineers are designing a system by which a falling mass \(m\) imparts kinetic energy to a rotating uniform drum to which it is attached by thin, very light wire wrapped around the rim of the drum (Fig. P9.68). There is no appreciable friction in the axle of the drum, and everything starts from rest. This system is being tested on earth, but it is to be used on Mars, where the acceleration due to gravity is 3.71 \(\mathrm{m} / \mathrm{s}^{2} .\) In the earth tests, when \(m\) is set to 15.0 \(\mathrm{kg}\) and allowed to fall through \(5.00 \mathrm{m},\) it gives 250.0 \(\mathrm{J}\) of kinetic energy to the drum. (a) If the system is operated on Mars, through what distance would the 15.0 -kg mass have to fall to give the same amount of kinetic energy to the drum? (b) How fast would the 15.0 -kg mass be moving on Mars just as the drum gained 250.0 \(\mathrm{J}\) of kinetic energy?

The motor of a table saw is rotating at 3450 rev/min. A pulley attached to the motor shaft drives a second pulley of half the diameter by moans of a V-belt. A circular saw blade of diameter 0.208 \(\mathrm{m}\) is mounted on the same rotating shaft as the second pulley. (a) The operator is careless and the blade catches and throws back a small piece of wood. This piece of wood moves with linear speed equal to the tangential speed of the rim of the blade. What is this speed? (b) Calculate the radial acceleration of points on the outer edge of the blade to see why sawdust doesn't to its teeth.
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