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Ch 10: Interactions and Potential Energy
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796당신이 사용하는 게 아니라요?교과서 변경
모든 교과서Knight Calc 5th EditionCh 10: Interactions and Potential Energy문제 41
10장, 문제 41

A 50 g mass is attached to a light, rigid, 75-cm-long rod. The other end of the rod is pivoted so that the mass can rotate in a vertical circle. What speed does the mass need at the bottom of the circle to barely make it over the top of the circle?

검증된 단계별 안내
1
Step 1: Identify the forces acting on the mass at the top of the circle. At the top, the gravitational force and the centripetal force must be balanced for the mass to barely make it over the top. The centripetal force is provided entirely by gravity in this case.
Step 2: Write the condition for the mass to barely make it over the top of the circle. The centripetal force required at the top is given by \( F_c = \frac{m v_{top}^2}{r} \), where \( m \) is the mass, \( v_{top} \) is the speed at the top, and \( r \) is the radius of the circle. For the mass to barely make it over, \( v_{top} \) must be such that \( F_c = mg \), where \( g \) is the acceleration due to gravity.
Step 3: Solve for \( v_{top} \). Using \( F_c = mg \), substitute \( mg = \frac{m v_{top}^2}{r} \). Cancel \( m \) from both sides to get \( v_{top}^2 = gr \). Take the square root to find \( v_{top} = \sqrt{gr} \).
Step 4: Use conservation of mechanical energy to relate the speed at the bottom of the circle \( v_{bottom} \) to the speed at the top \( v_{top} \). At the bottom, the mass has both kinetic energy and potential energy, while at the top, it has only potential energy. Write the energy conservation equation: \( KE_{bottom} + PE_{bottom} = KE_{top} + PE_{top} \). Substitute \( KE = \frac{1}{2}mv^2 \) and \( PE = mgh \), where \( h \) is the height relative to the bottom.
Step 5: Solve for \( v_{bottom} \). At the bottom, \( h = 0 \), so \( PE_{bottom} = 0 \). At the top, \( h = 2r \), so \( PE_{top} = mg(2r) \). Substitute \( KE_{bottom} = \frac{1}{2}m v_{bottom}^2 \), \( KE_{top} = \frac{1}{2}m v_{top}^2 \), and \( PE_{top} = mg(2r) \) into the energy conservation equation: \( \frac{1}{2}m v_{bottom}^2 = \frac{1}{2}m v_{top}^2 + mg(2r) \). Cancel \( m \) from all terms and solve for \( v_{bottom} \).

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
영상 길이:
6m

주요 개념

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

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. For an object in vertical circular motion, this force is provided by the gravitational force and the tension in the rod. At the top of the circle, the gravitational force must be sufficient to provide the necessary centripetal force to keep the mass in circular motion.
추천 영상:
가이드 코스
06:48
Intro to Centripetal Forces

Gravitational Potential Energy

Gravitational potential energy (GPE) is the energy an object possesses due to its position in a gravitational field. In the context of circular motion, as the mass rises to the top of the circle, its GPE increases while its kinetic energy decreases. To just make it over the top, the mass must have enough speed at the bottom to convert kinetic energy into GPE at the top.
추천 영상:
가이드 코스
06:35
Gravitational Potential Energy

Conservation of Energy

The principle of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. In this scenario, the total mechanical energy (kinetic plus potential) of the mass must remain constant throughout its motion. This means that the kinetic energy at the bottom of the circle must equal the sum of the potential energy at the top and any remaining kinetic energy needed to maintain circular motion.
추천 영상:
가이드 코스
06:24
Conservation Of Mechanical Energy
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