/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 91 A block with mass \(m\) is revol... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 91

A block with mass \(m\) is revolving with linear speed \(v_{1}\) in a circle of radius \(r_{1}\) on a frictionless horizontal surface (see Fig. E10.42). The string is slowly pulled from below until the radius of the circle in which the block is revolving is reduced to \(r_{2}\). (a) Calculate the tension \(T\) in the string as a function of \(r\), the distance of the block from the hole. Your answer will be in terms of the initial velocity \(v_{1}\) and the radius \(r_{1} .\) (b) Use \(W=\int_{r_{1}}^{r_{2}} \overrightarrow{\boldsymbol{T}}(r) \cdot d \overrightarrow{\boldsymbol{r}}\) to calculate the work done by \(\vec{T}\) when \(r\) changes from \(r_{1}\) to \(r_{2}\). (c) Compare the results of part (b) to the change in the kinetic energy of the block.

Short Answer

Expert verified
The tension in the string as a function of radius is \(T= m \cdot v_{1}^{2} \cdot r_{1}^{2}/r^{3}\). The work done by tension is \(W = \frac{1}{2} m \cdot v_{1}^{2} \cdot r_{1}^{2} [\frac{1}{r_{2}^{2}} - \frac{1}{r_{1}^{2}}]\). The change in kinetic energy is \( \Delta K = \frac{1}{2} m \cdot v_{1}^{2} [\frac{r_{1}^{2}}{r_{2}^{2}} - 1]\). The work done by the tension equals the change in kinetic energy of the block.

Step by step solution

01

Calculate tension

First, consider the block revolving in the circle. Here, the circular motion is maintained by the balance of two forces: the tension in the string \(T\) and the centripetal force due to circular motion, which is given by \(m \cdot v^{2}/r\). Thus, we can write that \(T = m \cdot v^{2}/r\). Now, according to the law of conservation of angular momentum, the initial angular momentum of the block should equal its final angular momentum. This means that \(m \cdot v_{1} \cdot r_{1} = m \cdot v \cdot r\). Solving for \(v\), we have \(v = v_{1} \cdot r_{1}/r\). Substituting this into the tension equation we find: \(T = m \cdot (v_{1} \cdot r_{1}/r)^{2}/r = m \cdot v_{1}^{2} \cdot r_{1}^{2}/r^{3}\).
02

Calculate work done

The work done by the tension from \(r_{1}\) to \(r_{2}\) is computed using a definite integral: \(W = \int_{r_{1}}^{r_{2}} T \cdot dr = \int_{r_{1}}^{r_{2}} m \cdot v_{1}^{2} \cdot r_{1}^{2}/r^{3} dr\). This integral can be simplified to: \(W = m \cdot v_{1}^{2} \cdot r_{1}^{2} \int_{r_{1}}^{r_{2}} 1/r^{3} dr = -m \cdot v_{1}^{2} \cdot r_{1}^{2} [\frac{1}{2r^{2}}]_{r_{1}}^{r_{2}}\). Evaluating the definite integral gives \(W = \frac{1}{2} m \cdot v_{1}^{2} \cdot r_{1}^{2} [\frac{1}{r_{2}^{2}} - \frac{1}{r_{1}^{2}}]\).
03

Kinetic energy comparison

The difference in kinetic energy \(K_{2} - K_{1}\) from radius \(r_{1}\) to \(r_{2}\) is evaluated by calculating the kinetic energy at each radius. The kinetic energy at the radius \(r\) is \(K = \frac{1}{2} m \cdot v^{2} = \frac{1}{2} m \cdot (v_{1} \cdot r_{1}/r)^{2} = \frac{1}{2} m \cdot v_{1}^{2} \cdot r_{1}^{2}/r^{2}\). Evaluating at \(r_{1}\) and \(r_{2}\) gives \(K_{1} = \frac{1}{2} m \cdot v_{1}^{2}\), \(K_{2} = \frac{1}{2} m \cdot v_{1}^{2} \cdot r_{1}^{2}/r_{2}^{2}\). The difference is \(K_{2} - K_{1} = \frac{1}{2} m \cdot v_{1}^{2} [\frac{r_{1}^{2}}{r_{2}^{2}} - 1]\), which equal to work done by tension.

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Key Concepts

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

Conservation of Angular Momentum
The principle of conservation of angular momentum is pivotal in understanding circular motion. It states that if no external torque acts on a system, the total angular momentum of the system remains constant over time. In the given exercise where a block revolves on a frictionless surface, this principle is crucial.

As the block's path radius decreases from an initial radius of \(r_{1}\) to a final radius of \(r_{2}\), we see that no external forces are causing angular acceleration. This allows us to assert that the block's initial and final angular momenta are equal. Mathematically, this is represented by \(m \cdot v_{1} \cdot r_{1} = m \cdot v \cdot r\), where \(v_{1}\) is the initial velocity, \(v\) is the final velocity, and \(m\) is the mass of the block.

Understanding this concept is critical in calculating both the tension in the string and the block's final velocity, as shown in the solution. It provides a foundation for exploring circular motion dynamics in objects ranging from small blocks to celestial bodies.
Centripetal Force
Centripetal force plays an essential role in keeping an object moving in a circular path. It can be thought of as the 'center-seeking' force that keeps an object on its curved trajectory. In our exercise, this force is provided by the tension in the string, which ensures that the block continues to move in a circular path.

When the block revolves around a circle with radius \(r\), the centripetal force needed for this motion equals \(m \cdot v^{2}/r\). The formula encapsulates how the force depends on the mass of the object (\(m\)), the velocity (\(v\)), and the radius (\(r\)) of the circular path. This is why when the string is pulled, affecting the radius, the velocity of the block must also change to maintain the balance of forces.

In the context of our problem, understanding the relationship between the centripetal force and the tension helps in explaining the system's behavior as the radius changes. It bridges the gap between abstract concepts and real-world phenomena, such as vehicles turning corners or planets orbiting stars.
Work-Energy Theorem
The work-energy theorem is a fundamental concept in physics that relates the work done on an object to its change in kinetic energy. When a force, in this case the tension in the string, does work on the block, it changes the block's kinetic energy.

In our exercise, the work done by the tension in the string as the radius changes from \(r_{1}\) to \(r_{2}\) is calculated using the integral \(W=\int_{r_{1}}^{r_{2}} T \cdot dr\), which eventually equates to the change in the block's kinetic energy. This demonstrates that the work done by the tension is not lost but is rather converted into kinetic energy, thereby increasing or decreasing the block鈥檚 speed.

Understanding how work leads to a change in kinetic energy reinforces the conservation of energy in physical systems. This principle not only answers how objects move but also offers insights into the efficiencies of machines and the behaviors of dynamic systems in various fields of science and engineering.

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Most popular questions from this chapter

A \(5.00 \mathrm{~kg}\) ball is dropped from a height of \(12.0 \mathrm{~m}\) above one end of a uniform bar that pivots at its center. The bar has mass 8.00 \(\mathrm{kg}\) and is \(4.00 \mathrm{~m}\) in length. At the other end of the bar sits another 5.00 \(\mathrm{kg}\) ball, unattached to the bar. The dropped ball sticks to the bar after the collision. How high will the other ball go after the collision?

You live on a planet far from ours. Based on extensive communication with a physicist on earth, you have determined that all laws of physics on your planet are the same as ours and you have adopted the same units of seconds and meters as on earth. But you suspect that the value of \(g\), the acceleration of an object in free fall near the surface of your planet, is different from what it is on earth. To test this, you take a solid uniform cylinder and let it roll down an incline. The vertical height \(h\) of the top of the incline above the lower end of the incline can be varied. You measure the speed \(v_{\mathrm{cm}}\) of the center of mass of the cylinder when it reaches the bottom for various values of \(h .\) You plot \(v_{\mathrm{cm}}^{2}\) (in \(\mathrm{m}^{2} / \mathrm{s}^{2}\) ) versus \(h\) (in \(\mathrm{m}\) ) and find that your data lie close to a straight line with a slope of \(6.42 \mathrm{~m} / \mathrm{s}^{2}\). What is the value of \(g\) on your planet?

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 static 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 \(2.20 \mathrm{~kg}\) hoop \(1.20 \mathrm{~m}\) in diameter is rolling to the right without slipping on a horizontal floor at a steady \(2.60 \mathrm{rad} / \mathrm{s}\). (a) How fast is its center moving? (b) What is the total kinetic energy of the hoop? (c) Find the velocity vector of each of the following points, as viewed by a person at rest on the ground: (i) the highest point on the hoop; (ii) the lowest point on the hoop; (iii) a point on the right side of the hoop, midway between the top and the bottom. (d) Find the velocity vector for each of the points in part (c), but this time as viewed by someone moving along with the same velocity as the hoop.

A block with mass \(m=5.00 \mathrm{~kg}\) slides down a surface inclined \(36.9^{\circ}\) to the horizontal (Fig. \(\mathbf{P 1 0 . 6 6 )}\). The coefficient of kinetic friction is \(0.25 .\) A string attached to the block is wrapped around a flywheel on a fixed axis at \(O .\) The flywheel has mass \(25.0 \mathrm{~kg}\) and moment of inertia \(0.500 \mathrm{~kg} \cdot \mathrm{m}^{2}\) with respect to the axis of rotation. The string pulls without slipping at a perpendicular distance of \(0.200 \mathrm{~m}\) from that axis. (a) What is the acceleration of the block down the plane? (b) What is the tension in the string?
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