NEET Exam > NEET Questions > A wheel has a constant angular acceleration o...
Start Learning for Free
A wheel has a constant angular acceleration of 2 rad/second square, during a certain 4.0s interval ,it turns through an angle of 72 rad . assuming that at t-0, angular speed -6 rad /sec how long,is motion at the start of this 4.0 second interval ?
Most Upvoted Answer
A wheel has a constant angular acceleration of 2 rad/second square, du...
Given data:
Angular acceleration (α) = 2 rad/s²
Angle turned (θ) = 72 rad
Initial angular speed (ω₀) = -6 rad/s
Time interval (t) = 4.0 s
Calculating the time at the start of the interval:
To find the time at the start of the interval, we need to use the kinematic equation for rotational motion:
θ = ω₀t + (1/2)αt²
Step 1: Calculate the final angular speed (ω):
Using the formula:
ω = ω₀ + αt
Substituting the given values:
ω = -6 + 2(4.0) = -6 + 8 = 2 rad/s
Step 2: Rearrange the kinematic equation:
θ = ω₀t + (1/2)αt²
Rearranging the equation, we get:
(1/2)αt² + ω₀t - θ = 0
Step 3: Solve the quadratic equation:
Using the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
Where:
a = (1/2)α = α/2
b = ω₀
c = -θ
Substituting the values, we get:
t = (-(ω₀) ± √((ω₀)² - 4(a)(c))) / (2(a))
Plugging in the values:
t = (-(2) ± √((2)² - 4(α/2)(-72))) / (2(α/2))
Simplifying further:
t = (-2 ± √(4 + 8α)) / α
Step 4: Calculate the time at the start of the interval:
Using the positive value of the quadratic equation to represent time, we have:
t = (-2 + √(4 + 8α)) / α
Substituting the given value of α:
t = (-2 + √(4 + 8(2))) / 2
Simplifying further:
t = (-2 + √(4 + 16)) / 2
t = (-2 + √20) / 2
t = (-2 + 2√5) / 2
t = -1 + √5
Hence, the motion at the start of this 4.0 second interval is approximately equal to -1 + √5 seconds.
Angular acceleration (α) = 2 rad/s²
Angle turned (θ) = 72 rad
Initial angular speed (ω₀) = -6 rad/s
Time interval (t) = 4.0 s
Calculating the time at the start of the interval:
To find the time at the start of the interval, we need to use the kinematic equation for rotational motion:
θ = ω₀t + (1/2)αt²
Step 1: Calculate the final angular speed (ω):
Using the formula:
ω = ω₀ + αt
Substituting the given values:
ω = -6 + 2(4.0) = -6 + 8 = 2 rad/s
Step 2: Rearrange the kinematic equation:
θ = ω₀t + (1/2)αt²
Rearranging the equation, we get:
(1/2)αt² + ω₀t - θ = 0
Step 3: Solve the quadratic equation:
Using the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
Where:
a = (1/2)α = α/2
b = ω₀
c = -θ
Substituting the values, we get:
t = (-(ω₀) ± √((ω₀)² - 4(a)(c))) / (2(a))
Plugging in the values:
t = (-(2) ± √((2)² - 4(α/2)(-72))) / (2(α/2))
Simplifying further:
t = (-2 ± √(4 + 8α)) / α
Step 4: Calculate the time at the start of the interval:
Using the positive value of the quadratic equation to represent time, we have:
t = (-2 + √(4 + 8α)) / α
Substituting the given value of α:
t = (-2 + √(4 + 8(2))) / 2
Simplifying further:
t = (-2 + √(4 + 16)) / 2
t = (-2 + √20) / 2
t = (-2 + 2√5) / 2
t = -1 + √5
Hence, the motion at the start of this 4.0 second interval is approximately equal to -1 + √5 seconds.
Community Answer
A wheel has a constant angular acceleration of 2 rad/second square, du...
Not understanding
Attention NEET Students!
To make sure you are not studying endlessly, EduRev has designed NEET study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in NEET.
|
Explore Courses for NEET exam
|
|
Similar NEET Doubts
Top Courses for NEETView all
A wheel has a constant angular acceleration of 2 rad/second square, during a certain 4.0s interval ,it turns through an angle of 72 rad . assuming that at t-0, angular speed -6 rad /sec how long,is motion at the start of this 4.0 second interval ?
Question Description
A wheel has a constant angular acceleration of 2 rad/second square, during a certain 4.0s interval ,it turns through an angle of 72 rad . assuming that at t-0, angular speed -6 rad /sec how long,is motion at the start of this 4.0 second interval ? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A wheel has a constant angular acceleration of 2 rad/second square, during a certain 4.0s interval ,it turns through an angle of 72 rad . assuming that at t-0, angular speed -6 rad /sec how long,is motion at the start of this 4.0 second interval ? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A wheel has a constant angular acceleration of 2 rad/second square, during a certain 4.0s interval ,it turns through an angle of 72 rad . assuming that at t-0, angular speed -6 rad /sec how long,is motion at the start of this 4.0 second interval ?.
A wheel has a constant angular acceleration of 2 rad/second square, during a certain 4.0s interval ,it turns through an angle of 72 rad . assuming that at t-0, angular speed -6 rad /sec how long,is motion at the start of this 4.0 second interval ? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A wheel has a constant angular acceleration of 2 rad/second square, during a certain 4.0s interval ,it turns through an angle of 72 rad . assuming that at t-0, angular speed -6 rad /sec how long,is motion at the start of this 4.0 second interval ? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A wheel has a constant angular acceleration of 2 rad/second square, during a certain 4.0s interval ,it turns through an angle of 72 rad . assuming that at t-0, angular speed -6 rad /sec how long,is motion at the start of this 4.0 second interval ?.
Solutions for A wheel has a constant angular acceleration of 2 rad/second square, during a certain 4.0s interval ,it turns through an angle of 72 rad . assuming that at t-0, angular speed -6 rad /sec how long,is motion at the start of this 4.0 second interval ? in English & in Hindi are available as part of our courses for NEET.
Download more important topics, notes, lectures and mock test series for NEET Exam by signing up for free.
Here you can find the meaning of A wheel has a constant angular acceleration of 2 rad/second square, during a certain 4.0s interval ,it turns through an angle of 72 rad . assuming that at t-0, angular speed -6 rad /sec how long,is motion at the start of this 4.0 second interval ? defined & explained in the simplest way possible. Besides giving the explanation of
A wheel has a constant angular acceleration of 2 rad/second square, during a certain 4.0s interval ,it turns through an angle of 72 rad . assuming that at t-0, angular speed -6 rad /sec how long,is motion at the start of this 4.0 second interval ?, a detailed solution for A wheel has a constant angular acceleration of 2 rad/second square, during a certain 4.0s interval ,it turns through an angle of 72 rad . assuming that at t-0, angular speed -6 rad /sec how long,is motion at the start of this 4.0 second interval ? has been provided alongside types of A wheel has a constant angular acceleration of 2 rad/second square, during a certain 4.0s interval ,it turns through an angle of 72 rad . assuming that at t-0, angular speed -6 rad /sec how long,is motion at the start of this 4.0 second interval ? theory, EduRev gives you an
ample number of questions to practice A wheel has a constant angular acceleration of 2 rad/second square, during a certain 4.0s interval ,it turns through an angle of 72 rad . assuming that at t-0, angular speed -6 rad /sec how long,is motion at the start of this 4.0 second interval ? tests, examples and also practice NEET tests.
|
|
Explore Courses for NEET exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup