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Ch 08: Dynamics II: Motion in a Plane
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 8, Problem 69a

The 10 mg bead in FIGURE CP8.69 is free to slide on a frictionless wire loop. The loop rotates about a vertical axis with angular velocity ω. If ω is less than some critical value ω꜀, the bead sits at the bottom of the spinning loop. When ω > ω꜀, the bead moves out to some angle θ. What is ω꜀ in rpm for the loop shown in the figure?

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Step 1: Analyze the forces acting on the bead. When the bead is at the bottom of the loop, the forces acting on it are gravity (mg) and the normal force from the wire. As the loop rotates, a centrifugal force acts outward, which depends on the angular velocity ω.
Step 2: Determine the condition for the bead to remain at the bottom of the loop. The bead will stay at the bottom as long as the centrifugal force is less than the gravitational force. The centrifugal force is given by F_c = mω²R, where R is the radius of the loop.
Step 3: Set up the critical condition for ω꜀. At the critical angular velocity ω꜀, the centrifugal force equals the gravitational force. This gives the equation mω꜀²R = mg, where g is the acceleration due to gravity.
Step 4: Solve for ω꜀. Rearrange the equation to find ω꜀: ω꜀ = √(g/R). This gives the critical angular velocity in radians per second. To convert ω꜀ to revolutions per minute (rpm), use the conversion factor: 1 rad/s = (60 / 2π) rpm.
Step 5: Substitute the given values into the equation. Use the mass of the bead (10 mg = 0.01 g = 0.00001 kg), the radius of the loop (R, as provided in the figure or problem statement), and g = 9.8 m/s² to calculate ω꜀ in rpm.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Centripetal Force

Centripetal force is the net force required to keep an object moving in a circular path, directed towards the center of the circle. In this scenario, the bead experiences centripetal force due to its circular motion on the rotating loop. This force is essential for determining the conditions under which the bead remains at the bottom or moves outward as the loop's angular velocity changes.
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Angular Velocity (ω)

Angular velocity (ω) is a measure of how quickly an object rotates around an axis, expressed in radians per second or revolutions per minute (rpm). In this problem, ω influences the motion of the bead on the loop. Understanding how angular velocity affects the forces acting on the bead is crucial for determining the critical value (ω₀) at which the bead begins to move outward.
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Critical Angular Velocity (ω₀)

The critical angular velocity (ω₀) is the threshold value of angular velocity at which the dynamics of the system change, causing the bead to transition from a stable position at the bottom of the loop to an outward position at an angle θ. This concept is vital for solving the problem, as it involves analyzing the balance of forces acting on the bead at different angular velocities.
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Related Practice
Textbook Question

A 500 g steel block rotates on a steel table while attached to a 1.2-m-long hollow tube as shown in FIGURE CP8.70. Compressed air fed through the tube and ejected from a nozzle on the back of the block exerts a thrust force of 4.0 N perpendicular to the tube. The maximum tension the tube can withstand without breaking is 50 N. If the block starts from rest, how many revolutions does it make before the tube breaks?

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Textbook Question

In the absence of air resistance, a projectile that lands at the elevation from which it was launched achieves maximum range when launched at a 45° angle. Suppose a projectile of mass m is launched with speed into a headwind that exerts a constant, horizontal retarding force Fwind=Fwindı^.\(\vec{F}\)_{\(\text{wind}\)} = -F_{\(\text{wind}\)} \(\hat{\imath}\). By what percentage is the maximum range of a 0.50 kg ball reduced if Fwind=0.60 NF_{\(\text{wind}\)}=0.60\(\text{ N}\)?

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Textbook Question

For safety, elevators have a rotational governor, a device that is attached to and rotates with one of the elevator's pulleys. The governor, shown in FIGURE P8.63, is a disk with two hollow channels holding springs with metal blocks of mass m attached to their free ends. The faster the governor spins, the more the springs stretch. At a critical angular velocity ωc, the metal blocks contact the housing, which completes a circuit and activates an emergency brake. The spring force on a mass, which we will explore more thoroughly in Chapter 9, is FSp = k(r - L), where k is the spring constant measured in N/m, and L is the relaxed (unstretched) length of the spring. Suppose a rotational governor has L = 0.80R and the emergency brake activates when the metal blocks reach r = R. What is the critical angular velocity in rpm if R = 15cm, k = 20 N/m, and m = 25g? Ignore gravity.

Textbook Question

If a vertical cylinder of water (or any other liquid) rotates about its axis, as shown in FIGURE CP8.72, the surface forms a smooth curve. Assuming that the water rotates as a unit (i.e., all the water rotates with the same angular velocity), show that the shape of the surface is a parabola described by the equation z = (ω2 / 2g) r2. Hint: Each particle of water on the surface is subject to only two forces: gravity and the normal force due to the water underneath it. The normal force, as always, acts perpendicular to the surface.

Textbook Question

In the absence of air resistance, a projectile that lands at the elevation from which it was launched achieves maximum range when launched at a 45° angle. Suppose a projectile of mass m is launched with speed into a headwind that exerts a constant, horizontal retarding force Fwind=Fwindı^.\(\vec{F}\)_{\(\text{wind}\)} = -F_{\(\text{wind}\)} \(\hat{\imath}\).. Find an expression for the angle at which the range is maximum.

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