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A small sphere with mass \(m\) is attached to a massless rod of length \(L\) that is pivoted at the top, forming a simple pendulum. The pendulum is pulled to one side so that the rod is at an angle \(\theta\) from the vertical, and released from rest. (a) In a diagram, show the pendulum just after it is released. Draw vectors representing the forces acting on the small sphere and the acceleration of the sphere. Accuracy counts! At this point, what is the linear acceleration of the sphere? (b) Repeat part (a) for the instant when the pendulum rod is at an angle \(\theta / 2\) from the vertical. (c) Repeat part (a) for the instant when the pendulum rod is vertical. At this point, what is the linear speed of the sphere?

Short Answer

Expert verified
The linear acceleration of the sphere just after release is \(gsin(\theta)\) and at an angle \(\theta/2\) from the vertical, it is \(gsin(\theta/2)\). When the pendulum rod is vertical, the linear speed of the sphere is \(\sqrt{2gL(1-cos(\theta))}\). There is no linear acceleration at this point since the pendulum is at its lowest point.

Step by step solution

01

- Initial Pendulum Release

The initial situation is where the pendulum is just released. Here, we have two forces acting on it - the gravitational force (\(mg\)) and the tension in the rod. We need to find the component of gravitation force in the direction of motion which is \(mgsin(\theta)\). Our diagram will show vectors representing these two forces along with the direction of the sphere’s acceleration which is in the same direction as \(mgsin(\theta)\). The linear acceleration is obtained by equating the net force to mass times acceleration using Newton's second law, \(F=ma\). So we get \( a = gsin(\theta)\).
02

- Pendulum at Half Initial Angle

When the pendulum is at an angle \(\theta/2\) from the vertical, the forces acting on the sphere and the direction of acceleration remain the same, except now the angle has halved. The net force changes and the linear acceleration of the sphere becomes \(a = gsin(\theta/2)\). So the linear acceleration decreases compared to the initial release.
03

- Pendulum's Lowest Point

The pendulum's rod becomes vertical at the lowest point. Here, the direction of the gravitational force vector changes to be directly downwards, but there is no component in the direction of motion since it's at the lowest point of its path. This means there is no net force acting in the direction of the pendulum's motion, therefore acceleration is zero. To find the linear speed of the sphere at this point, we use the principle of conservation of mechanical energy. As the pendulum swings down, the gravitational potential energy is converted into kinetic energy. From the energy conservation equation, \(\frac{1}{2}mv^2 = mgh\), where \(h = L - Lcos(\theta) = L(1-cos(\theta))\), we solve for velocity to get \(v=\sqrt{2gL(1-cos(\theta))}\), which gives the linear speed at the lowest point.

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Key Concepts

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

Gravitational Force
The gravitational force is a fundamental interaction that acts on objects with mass, attracting them towards each other. In the context of a simple pendulum, the gravitational force is what pulls the sphere downward towards the Earth's center. This force can be represented by the equation \( F_{grav} = mg \) where \( m \) is the mass of the sphere and \( g \) is the acceleration due to gravity, approximately \( 9.81 \, m/s^2 \) near the Earth’s surface.

When a pendulum is pulled aside and released, the gravitational force acts vertically downward. However, only the component of this force along the arc of the pendulum's swing, \( F = mgsin(\theta) \) contributes to the sphere's acceleration. It's this force that propels the pendulum in motion, transferring potential energy into kinetic energy as it accelerates towards its lowest point.
Newton's Second Law
Newton's second law of motion outlines the relationship between an object’s mass \( m \) , its acceleration \( a \) , and the net force \( F \) acting on it, generally stated as \( F = ma \).

For the simple pendulum, we apply this law to find the sphere's acceleration just after release. Assuming the only force working in the direction of the pendulum's motion is the gravitational component \( mgsin(\theta) \) and neglecting air resistance, the linear acceleration \( a \) at angle \( \theta \) can be calculated using \( a = gsin(\theta) \). This equation signifies that the pendulum's acceleration is proportional to the sine of the angle and gravity, reflecting how the pendulum moves faster as it's released from a larger angle.
Conservation of Mechanical Energy
The conservation of mechanical energy principle states that in a system with only conservative forces (like gravity), the total mechanical energy (sum of potential and kinetic energy) remains constant if no energy is added or removed from the system.

In our pendulum problem, as the pendulum swings downward from its initial raised position, potential energy is converted into kinetic energy. The total mechanical energy at the starting point (potential energy) is equal to the mechanical energy at the pendulum's lowest point (kinetic energy). By applying the conservation of energy, we can express this as \( \frac{1}{2}mv^2 = mgh \), where \( h \) is the height the pendulum falls. Solving for velocity gives us the linear speed of the sphere at the lowest point, illustrating how the pendulum's motion is a continual exchange between potential and kinetic energy.
Pendulum Linear Acceleration
Linear acceleration refers to the rate of change of linear speed. In a simple pendulum, the linear acceleration of the sphere depends on the gravitational force's component along the swing's arc. Immediately after the pendulum is released, the sphere accelerates as it begins to move along the arc.

At the initial angle \( \theta \) and when the pendulum makes an angle \( \theta/2 \) with the vertical, the linear acceleration is \( gsin(\theta) \) and \( gsin(\theta/2) \) respectively. The linear acceleration is greatest at the start and decreases as the pendulum approaches the vertical, due to the decrease in the gravitational force component along the swing. It is essential to recognize that the linear acceleration is not constant throughout the swing but varies with the cosine of the displacement angle.
Pendulum Linear Speed
Pendulum linear speed is the rate at which the sphere moves along its circular path. This speed is zero just as the pendulum is released and increases as it falls due to the component of gravitational force pulling it along its path.

At any point in its swing, we can calculate the pendulum’s linear speed by considering the conservation of mechanical energy. At the lowest point of its swing, all the pendulum's potential energy has been converted into kinetic energy. Using \( v = \sqrt{2gL(1-cos(\theta))} \) we find the linear speed at this point. It is important to note that this speed is the maximum the sphere reaches and occurs only when the pendulum is at the bottom of its swing, at which point the speed then starts to decrease again as the pendulum ascends.

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Most popular questions from this chapter

BIO Weighing a Virus. In February 2004, scientists at Purdue University used a highly sensitive technique to measure the mass of a vaccinia virus (the kind used in smallpox vaccine). The procedure involved measuring the frequency of oscillation of a tiny sliver of silicon (just \(30 \mathrm{nm}\) long) with a laser, first without the virus and then after the virus had attached itself to the silicon. The difference in mass caused a change in the frequency. We can model such a process as a mass on a spring. (a) Show that the ratio of the frequency with the virus attached \(\left(f_{\mathrm{S}+\mathrm{V}}\right)\) to the frequency without the virus \(\left(f_{\mathrm{S}}\right)\) is given by \(f_{\mathrm{S}+\mathrm{V}} / f_{\mathrm{S}}=1 / \sqrt{1+\left(m_{\mathrm{V}} / m_{\mathrm{S}}\right)},\) where \(m_{\mathrm{V}}\) is the mass of the virus and \(m_{\mathrm{S}}\) is the mass of the silicon sliver. Notice that it is not necessary to know or measure the force constant of the spring. (b) In some data, the silicon sliver has a mass of \(2.10 \times 10^{-16} \mathrm{~g}\) and a frequency of \(2.00 \times 10^{15} \mathrm{~Hz}\) without the virus and \(2.87 \times 10^{14} \mathrm{~Hz}\) with the virus. What is the mass of the virus, in grams and in femtograms?

Two uniform solid spheres, each with mass \(M=0.800 \mathrm{~kg}\) and radius \(R=0.0800 \mathrm{~m},\) are connected by a short, light rod that is along a diameter of each sphere and are at rest on a horizontal tabletop. A spring with force constant \(k=160 \mathrm{~N} / \mathrm{m}\) has one end attached to the wall and the other end attached to a friction less ring that passes over the rod at the center of mass of the spheres, which is midway between the centers of the two spheres. The spheres are each pulled the same distance from the wall, stretching the spring, and released. There is sufficient friction between the tabletop and the spheres for the spheres to roll without slipping as they move back and forth on the end of the spring. Show that the motion of the center of mass of the spheres is simple harmonic and calculate the period.

A \(40.0 \mathrm{~N}\) force stretches a vertical spring \(0.250 \mathrm{~m}\). (a) What mass must be suspended from the spring so that the system will oscillate with a period of \(1.00 \mathrm{~s} ?\) (b) If the amplitude of the motion is \(0.050 \mathrm{~m}\) and the period is that specified in part (a), where is the object and in what direction is it moving \(0.35 \mathrm{~s}\) after it has passed the equilibrium position, moving downward? (c) What force (magnitude and direction) does the spring exert on the object when it is \(0.030 \mathrm{~m}\) below the equilibrium position, moving upward?

Consider the system of two blocks and a spring shown in Fig. \(\mathrm{P} 14.66 .\) The horizontal surface is friction less, but there is static friction between the two blocks. The spring has force constant \(k=150 \mathrm{~N} / \mathrm{m} .\) The masses of the two blocks are \(m=0.500 \mathrm{~kg}\) and \(M=4.00 \mathrm{~kg} .\) You set the blocks into motion by releasing block \(M\) with the spring stretched a distance \(d\) from equilibrium. You start with small values of \(d,\) and then repeat with successively larger values. For small values of \(d,\) the blocks move together in SHM. But for larger values of \(d\) the top block slips relative to the bottom block when the bottom block is released. (a) What is the period of the motion of the two blocks when \(d\) is small enough to have no slipping? (b) The largest value \(d\) can have and there be no slipping is \(d=8.8 \mathrm{~cm} .\) What is the coefficient of static friction \(\mu_{\mathrm{s}}\) between the surfaces of the two blocks?

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