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Ch 12: Rotation of a Rigid Body
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 12, Problem 86

The two blocks in FIGURE CP12.86 are connected by a massless rope that passes over a pulley. The pulley is 12 cm in diameter and has a mass of 2.0 kg. As the pulley turns, friction at the axle exerts a torque of magnitude 0.50 N m. If the blocks are released from rest, how long does it take the 4.0 kg block to reach the floor?

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Identify the forces acting on the system: The two blocks are connected by a rope over a pulley. The 4.0 kg block experiences a downward gravitational force \( F_g = m g \), while the other block (mass unknown) experiences a similar force. The pulley has rotational inertia, and there is a torque due to friction at the axle.
Write the equations of motion for the blocks: For the 4.0 kg block, the net force is \( F_{net} = m_1 g - T_1 \), where \( T_1 \) is the tension in the rope on the side of the 4.0 kg block. For the other block, the net force is \( F_{net} = T_2 - m_2 g \), where \( T_2 \) is the tension on the other side of the rope.
Account for the rotational motion of the pulley: The pulley has a rotational inertia \( I = \frac{1}{2} M R^2 \), where \( M \) is the mass of the pulley and \( R \) is its radius. The net torque on the pulley is \( \tau_{net} = T_1 R - T_2 R - \tau_{friction} \), where \( \tau_{friction} = 0.50 \; \text{N·m} \). Use Newton's second law for rotation: \( \tau_{net} = I \alpha \), where \( \alpha \) is the angular acceleration.
Relate the linear acceleration of the blocks to the angular acceleration of the pulley: The linear acceleration \( a \) of the blocks is related to the angular acceleration \( \alpha \) of the pulley by \( a = \alpha R \). Substitute this relationship into the equations of motion for the blocks and the pulley.
Combine the equations to solve for the acceleration \( a \): Use the equations for the forces on the blocks and the torque on the pulley to form a system of equations. Solve for \( a \), the linear acceleration of the blocks. Once \( a \) is known, use the kinematic equation \( y = \frac{1}{2} a t^2 \) to solve for the time \( t \) it takes for the 4.0 kg block to reach the floor, where \( y \) is the distance the block falls.

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Key Concepts

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

Newton's Second Law of Motion

Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This principle is crucial for analyzing the motion of the blocks, as it allows us to calculate the forces acting on each block and determine their acceleration as they move under the influence of gravity and tension in the rope.
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Torque and Rotational Motion

Torque is a measure of the rotational force applied to an object, calculated as the product of the force and the distance from the pivot point. In this scenario, the torque exerted by friction at the axle of the pulley affects its angular acceleration, which in turn influences the linear acceleration of the blocks connected by the rope. Understanding torque is essential for analyzing the dynamics of the pulley system.
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Kinematic Equations

Kinematic equations describe the motion of objects under constant acceleration. These equations relate displacement, initial velocity, final velocity, acceleration, and time. In this problem, we can use these equations to determine how long it takes for the 4.0 kg block to fall to the floor, given its initial conditions and the calculated acceleration from the forces acting on the system.
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Related Practice
Textbook Question

A rod of length L and mass M has a nonuniform mass distribution. The linear mass density (mass per length) is λ = cx2, where x is measured from the center of the rod and c is a constant. Find an expression for c in terms of L and M.

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Textbook Question

The bunchberry flower has the fastest-moving parts ever observed in a plant. Initially, the stamens are held by the petals in a bent position, storing elastic energy like a coiled spring. When the petals release, the tips of the stamen act like medieval catapults, flipping through a 60° angle in just 0.30 ms to launch pollen from anther sacs at their ends. The human eye just sees a burst of pollen; only high-speed photography reveals the details. As FIGURE CP12.91 shows, we can model the stamen tip as a 1.0-mm-long, 10 μg rigid rod with a 10 μg anther sac at the end. Although oversimplifying, we'll assume a constant angular acceleration. What is the speed of the anther sac as it releases its pollen?

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Textbook Question

A rod of length L and mass M has a nonuniform mass distribution. The linear mass density (mass per length) is λ = cx2 , where x is measured from the center of the rod and c is a constant. Find an expression in terms of L and M for the moment of inertia of the rod for rotation about an axis through the center.

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Textbook Question

During most of its lifetime, a star maintains an equilibrium size in which the inward force of gravity on each atom is balanced by an outward pressure force due to the heat of the nuclear reactions in the core. But after all the hydrogen 'fuel' is consumed by nuclear fusion, the pressure force drops and the star undergoes a gravitational collapse until it becomes a neutron star. In a neutron star, the electrons and protons of the atoms are squeezed together by gravity until they fuse into neutrons. Neutron stars spin very rapidly and emit intense pulses of radio and light waves, one pulse per rotation. These 'pulsing stars' were discovered in the 1960s and are called pulsars. a. A star with the mass (M = 2.0 X 1030 kg) and size (R = 7.0 x 108 m) of our sun rotates once every 30 days. After undergoing gravitational collapse, the star forms a pulsar that is observed by astronomers to emit radio pulses every 0.10 s. By treating the neutron star as a solid sphere, deduce its radius.

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Textbook Question

FIGURE P12.82 shows a cube of mass m sliding without friction at speed v0. It undergoes a perfectly elastic collision with the bottom tip of a rod of length d and mass M = 2m. The rod is pivoted about a frictionless axle through its center, and initially it hangs straight down and is at rest. What is the cube's velocity—both speed and direction—after the collision?

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Textbook Question

Objects that rotate in air or water experience a torque due to drag. With quadratic drag, a drag torque that's negligible at low rpm quickly becomes significant as the rpm increases. Consider a square bar with cross section a x a and length L. It is rotating on an axle through its center at angular velocity ω in a fluid of density ρ. Assume that the drag coefficient C𝒹 is constant along the length of the bar. Find an expression for the magnitude of the drag torque on the bar. Hint: Begin by considering the drag force on a small piece of the bar of length dr at distance r from the axle.

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