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Ch 10: Dynamics of Rotational Motion
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610당신이 사용하는 게 아니라요?교과서 변경
10장, 문제 40c

A small block on a frictionless, horizontal surface has a mass of 0.0250 kg. It is attached to a massless cord passing through a hole in the surface (Fig. E10.40). The block is originally revolving at a distance of 0.300 m from the hole with an angular speed of 2.85 rad/s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.150 m. Model the block as a particle. Find the change in kinetic energy of the block.

검증된 단계별 안내
1
Identify the initial and final conditions of the system. Initially, the block is revolving at a radius of 0.300 m with an angular speed of 2.85 rad/s. The final radius is 0.150 m.
Use the conservation of angular momentum to find the final angular speed. The initial angular momentum (L_i) is given by L_i = m * r_i^2 * ω_i, where m is the mass, r_i is the initial radius, and ω_i is the initial angular speed. The final angular momentum (L_f) is L_f = m * r_f^2 * ω_f, where r_f is the final radius and ω_f is the final angular speed. Set L_i = L_f to solve for ω_f.
Calculate the initial kinetic energy (KE_i) using the formula KE_i = 0.5 * m * (r_i * ω_i)^2.
Calculate the final kinetic energy (KE_f) using the formula KE_f = 0.5 * m * (r_f * ω_f)^2, where ω_f is obtained from the conservation of angular momentum.
Determine the change in kinetic energy by finding the difference ΔKE = KE_f - KE_i. This will give you the change in kinetic energy of the block as the radius changes.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Conservation of Angular Momentum

Angular momentum is conserved in a system where no external torques are acting. For a particle revolving in a circle, the angular momentum is given by L = mvr, where m is mass, v is tangential velocity, and r is the radius. As the radius changes, the velocity must adjust to keep the angular momentum constant, which is crucial for solving this problem.
추천 영상:
12:12
Conservation of Angular Momentum

Kinetic Energy in Rotational Motion

Kinetic energy in rotational motion is given by KE = 0.5 * m * v^2, where m is mass and v is tangential velocity. As the radius of the circle changes, the velocity changes due to conservation of angular momentum, affecting the kinetic energy. Understanding how kinetic energy depends on velocity is essential for calculating the change in kinetic energy.
추천 영상:
06:07
Intro to Rotational Kinetic Energy

Relationship Between Linear and Angular Velocity

The relationship between linear velocity (v) and angular velocity (ω) is v = ωr, where r is the radius of the circle. As the radius changes, the angular velocity remains constant, but the linear velocity changes. This relationship helps in determining how the velocity changes when the radius is altered, impacting the kinetic energy calculation.
추천 영상:
11:10
Converting Between Linear & Rotational
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교과서 질문

A small block on a frictionless, horizontal surface has a mass of 0.0250 kg. It is attached to a massless cord passing through a hole in the surface (Fig. E10.40). The block is originally revolving at a distance of 0.300 m from the hole with an angular speed of 2.85 rad/s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.150 m. Model the block as a particle. How much work was done in pulling the cord?

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교과서 질문

A small block on a frictionless, horizontal surface has a mass of 0.0250 kg. It is attached to a massless cord passing through a hole in the surface (Fig. E10.40). The block is originally revolving at a distance of 0.300 m from the hole with an angular speed of 2.85 rad/s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.150 m. Model the block as a particle. What is the new angular speed?

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A small block on a frictionless, horizontal surface has a mass of 0.0250 kg. It is attached to a massless cord passing through a hole in the surface (Fig. E10.40). The block is originally revolving at a distance of 0.300 m from the hole with an angular speed of 2.85 rad/s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.150 m. Model the block as a particle. Is the angular momentum of the block conserved? Why or why not?

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