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

Knight Calc 5th Edition

Ch 10: Interactions and Potential Energy

문제 44

10장, 문제 44

A block of mass m slides down a frictionless track, then around the inside of a circular loop-the-loop of radius R . From what minimum height h must the block start to make it around without falling off? Give your answer as a multiple of R.

검증된 단계별 안내

Step 1: Recognize that the block must have enough kinetic energy at the top of the loop to counteract the gravitational force and maintain contact with the track. At the top of the loop, the centripetal force is provided entirely by the gravitational force and the normal force (which can be zero at the minimum height).

Step 2: Write the condition for the block to just make it around the loop. At the top of the loop, the centripetal force is given by \( F_c = \frac{mv^2}{R} \), where \( v \) is the velocity of the block at the top of the loop. For the block to stay in contact, the gravitational force \( F_g = mg \) must equal or exceed \( F_c \). Thus, \( mg = \frac{mv^2}{R} \).

Step 3: Solve for the velocity \( v \) at the top of the loop. From \( mg = \frac{mv^2}{R} \), cancel \( m \) and rearrange to get \( v^2 = gR \). Therefore, \( v = \sqrt{gR} \).

Step 4: Use energy conservation to relate the initial height \( h \) to the velocity at the top of the loop. The total mechanical energy at the starting height is \( E = mgh \), and at the top of the loop, it is \( E = mgh_{top} + \frac{1}{2}mv^2 \), where \( h_{top} = 2R \) (the height of the loop). Equating the energies, \( mgh = mg(2R) + \frac{1}{2}m(\sqrt{gR})^2 \).

Step 5: Simplify the energy equation to solve for \( h \). Substitute \( v^2 = gR \) into the equation: \( mgh = mg(2R) + \frac{1}{2}mgR \). Cancel \( m \) and simplify: \( h = 2R + \frac{1}{2}R = \frac{5R}{2} \). Thus, the minimum height \( h \) is \( \frac{5R}{2} \).

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.

영상 길이:

주요 개념

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

Conservation of Energy

The principle of conservation of energy states that the total mechanical energy of an isolated system remains constant if only conservative forces are acting. In this scenario, the block's potential energy at height h is converted into kinetic energy as it descends the track. This relationship allows us to equate the initial potential energy to the kinetic energy required to maintain motion through the loop.

Centripetal Force

Centripetal force is the net force required to keep an object moving in a circular path and is directed towards the center of the circle. For the block to successfully navigate the loop without falling off, it must have sufficient speed at the top of the loop to provide the necessary centripetal force. This force is provided by the gravitational force acting on the block at that point.

Minimum Speed at the Top of the Loop

At the top of the loop, the block must have a minimum speed to ensure that the gravitational force is sufficient to provide the necessary centripetal force. This minimum speed can be derived from the condition that the gravitational force equals the required centripetal force at that point. If the speed is too low, the block will not have enough centripetal force to stay on the track.

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