Skip to main content
Ch 04: Kinematics in Two Dimensions
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 04: Kinematics in Two Dimensions문제 40b
4장, 문제 40b

A 5.0-m-diameter merry-go-round is initially turning with a 4.0 s period. It slows down and stops in 20 s. How many revolutions does the merry-go-round make as it stops?

검증된 단계별 안내
1
Step 1: Calculate the initial angular velocity (ω₀) of the merry-go-round using the formula ω₀ = 2π / T, where T is the period of rotation. Here, T = 4.0 s.
Step 2: Determine the angular acceleration (α) using the formula α = (ω - ω₀) / t, where ω is the final angular velocity (0 rad/s since it stops), ω₀ is the initial angular velocity, and t is the time it takes to stop (20 s).
Step 3: Use the kinematic equation for rotational motion θ = ω₀t + (1/2)αt² to calculate the total angular displacement (θ) in radians. Substitute the values of ω₀, α, and t into the equation.
Step 4: Convert the angular displacement θ from radians to revolutions using the relationship 1 revolution = 2π radians. Divide θ by 2π to find the number of revolutions.
Step 5: Summarize the process and ensure all units are consistent throughout the calculations to verify the result.

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
영상 길이:
7m
도움이 되었나요?

주요 개념

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

Angular Velocity

Angular velocity is a measure of how quickly an object rotates around an axis, typically expressed in radians per second. For a merry-go-round, it can be calculated from the period of rotation, which is the time taken for one complete revolution. The relationship is given by the formula ω = 2π/T, where ω is the angular velocity and T is the period.
추천 영상:
06:18
Intro to Angular Momentum

Angular Deceleration

Angular deceleration refers to the rate at which an object's angular velocity decreases over time. It is crucial for understanding how quickly the merry-go-round slows down. This can be calculated by dividing the change in angular velocity by the time taken to stop, allowing us to determine how many revolutions occur during the deceleration phase.
추천 영상:
12:12
Conservation of Angular Momentum

Total Revolutions

Total revolutions represent the cumulative number of complete turns an object makes during a specific time interval. For the merry-go-round, this can be calculated by integrating the angular velocity over the time it takes to stop. By knowing the initial angular velocity and the angular deceleration, one can find the total number of revolutions made as it comes to a halt.
추천 영상:
05:29
Total Internal Reflection
관련 실천
교과서 질문

Starting from rest, a DVD steadily accelerates to 500 rpm in 1.0 s, rotates at this angular speed for 3.0 s, then steadily decelerates to a halt in 2.0 s. How many revolutions does it make?

2078
views
교과서 질문

FIGURE EX4.36 shows the angular velocity graph of the crankshaft in a car. What is the crankshaft's angular acceleration at t = 3s

1667
views
교과서 질문

Your roommate is working on his bicycle and has the bike upside down. He spins the 60-cm-diameter wheel, and you notice that a pebble stuck in the tread goes by three times every second. What are the pebble's speed and acceleration?

2112
views
1
rank
교과서 질문

A speck of dust on a spinning DVD has a centripetal acceleration of 20 m/s3 . What would be the acceleration of the first speck of dust if the disk's angular velocity was doubled?

404
views
교과서 질문

A projectile's horizontal range over level ground is v02sin2θg\(\frac{v_0^2 \sin 2\theta}{g}\). At what launch angle or angles will the projectile land at half of its maximum possible range?

2391
views
교과서 질문

A spaceship maneuvering near Planet Zeta is located at r=(600i−400j+200k)×103km,\(\mathbf{r}\) = (600\(\mathbf{i}\) - 400\(\mathbf{j}\) + 200\(\mathbf{k}\)) \(\times\) 10^3 \, \(\text{km}\), relative to the planet, and traveling at v=9500im/s\(\mathbf{v}\) = 9500\(\mathbf{i}\) \, \(\text{m/s}\). It turns on its thruster engine and accelerates with a=(40i−20k)m/s2\(\mathbf{a}\) = (40\(\mathbf{i}\) - 20\(\mathbf{k}\)) \, \(\text{m/s}\)^2 for 35 min35\(\text{ min}\). What is the spaceship's position when the engine shuts off? Give your answer as a position vector measured in km\(\operatorname{km}\).

2136
views