Physics Exam > Physics Questions > A particle of mass 20 gm is moving in a circl...
Start Learning for Free
A particle of mass 20 gm is moving in a circle of 4cm radius with contant speed of 10 cm/sec. What is the angular momentum (gm. cm2 / sec) about a point on the axis of the circle 3 cm distance from the centre.
Correct answer is '1000'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
A particle of mass 20 gm is moving in a circle of 4cm radius with cont...
Angular momentum about a point on the axis of the circle
Since,
is perpendocular to the velocity
Since,
is perpendocular to the velocityMost Upvoted Answer
A particle of mass 20 gm is moving in a circle of 4cm radius with cont...
Given:
Mass of the particle, m = 20 gm = 0.02 kg
Radius of the circle, r = 4 cm = 0.04 m
Speed of the particle, v = 10 cm/sec = 0.1 m/sec
Distance of the point on the axis from the center, d = 3 cm = 0.03 m
To find:
Angular momentum of the particle about the given point.
Solution:
1. Angular Velocity:
The particle is moving in a circle with a constant speed, which means it has a constant tangential velocity. The tangential velocity is given by v = rω, where ω is the angular velocity.
Substituting the given values, we have:
0.1 m/sec = 0.04 m × ω
Simplifying the equation, we get:
ω = 0.1 m/sec / 0.04 m = 2.5 rad/sec
2. Angular Momentum:
The angular momentum of a particle about a point is given by L = Iω, where I is the moment of inertia and ω is the angular velocity.
The moment of inertia of a particle moving in a circle about an axis is given by I = mr^2.
Substituting the given values, we have:
I = 0.02 kg × (0.04 m)^2 = 0.02 kg × 0.0016 m^2 = 0.000032 kg·m^2
Substituting the values of I and ω, we have:
L = 0.000032 kg·m^2 × 2.5 rad/sec = 0.00008 kg·m^2·rad/sec
3. Converting to gm.cm^2/sec:
To convert the angular momentum to gm·cm^2/sec, we need to multiply by 1000.
L (in gm·cm^2/sec) = 0.00008 kg·m^2·rad/sec × 1000 gm/kg × 100 cm/m = 8 gm·cm^2/sec
4. Angular Momentum about the given point:
To calculate the angular momentum about the given point, we need to consider the perpendicular distance between the point and the axis of rotation.
The perpendicular distance, r', is given by r' = r - d = 0.04 m - 0.03 m = 0.01 m
Substituting the values of r' and ω, we have:
L = Iω = (mr'^2)ω = (0.02 kg × (0.01 m)^2) × 2.5 rad/sec = 0.02 kg × 0.0001 m^2 × 2.5 rad/sec = 0.00005 kg·m^2·rad/sec
Converting to gm·cm^2/sec, we have:
L (in gm·cm^2/sec) = 0.00005 kg·m^2·rad/sec × 1000 gm/kg × 100 cm/m = 5 gm·cm^2/sec
Therefore, the angular momentum of the particle about the given point is 5 gm·cm^2/sec.
Mass of the particle, m = 20 gm = 0.02 kg
Radius of the circle, r = 4 cm = 0.04 m
Speed of the particle, v = 10 cm/sec = 0.1 m/sec
Distance of the point on the axis from the center, d = 3 cm = 0.03 m
To find:
Angular momentum of the particle about the given point.
Solution:
1. Angular Velocity:
The particle is moving in a circle with a constant speed, which means it has a constant tangential velocity. The tangential velocity is given by v = rω, where ω is the angular velocity.
Substituting the given values, we have:
0.1 m/sec = 0.04 m × ω
Simplifying the equation, we get:
ω = 0.1 m/sec / 0.04 m = 2.5 rad/sec
2. Angular Momentum:
The angular momentum of a particle about a point is given by L = Iω, where I is the moment of inertia and ω is the angular velocity.
The moment of inertia of a particle moving in a circle about an axis is given by I = mr^2.
Substituting the given values, we have:
I = 0.02 kg × (0.04 m)^2 = 0.02 kg × 0.0016 m^2 = 0.000032 kg·m^2
Substituting the values of I and ω, we have:
L = 0.000032 kg·m^2 × 2.5 rad/sec = 0.00008 kg·m^2·rad/sec
3. Converting to gm.cm^2/sec:
To convert the angular momentum to gm·cm^2/sec, we need to multiply by 1000.
L (in gm·cm^2/sec) = 0.00008 kg·m^2·rad/sec × 1000 gm/kg × 100 cm/m = 8 gm·cm^2/sec
4. Angular Momentum about the given point:
To calculate the angular momentum about the given point, we need to consider the perpendicular distance between the point and the axis of rotation.
The perpendicular distance, r', is given by r' = r - d = 0.04 m - 0.03 m = 0.01 m
Substituting the values of r' and ω, we have:
L = Iω = (mr'^2)ω = (0.02 kg × (0.01 m)^2) × 2.5 rad/sec = 0.02 kg × 0.0001 m^2 × 2.5 rad/sec = 0.00005 kg·m^2·rad/sec
Converting to gm·cm^2/sec, we have:
L (in gm·cm^2/sec) = 0.00005 kg·m^2·rad/sec × 1000 gm/kg × 100 cm/m = 5 gm·cm^2/sec
Therefore, the angular momentum of the particle about the given point is 5 gm·cm^2/sec.
Free Test
FREE
| Start Free Test |
Community Answer
A particle of mass 20 gm is moving in a circle of 4cm radius with cont...
Angular momentum about a point on the axis of the circle
Since, is perpendocular to the velocity
Since,
is perpendocular to the velocity
|
Explore Courses for Physics exam
|
|
A particle of mass 20 gm is moving in a circle of 4cm radius with contant speed of 10 cm/sec. What is the angular momentum (gm. cm2 / sec) about a point on the axis of the circle 3 cm distance from the centre.Correct answer is '1000'. Can you explain this answer?
Question Description
A particle of mass 20 gm is moving in a circle of 4cm radius with contant speed of 10 cm/sec. What is the angular momentum (gm. cm2 / sec) about a point on the axis of the circle 3 cm distance from the centre.Correct answer is '1000'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A particle of mass 20 gm is moving in a circle of 4cm radius with contant speed of 10 cm/sec. What is the angular momentum (gm. cm2 / sec) about a point on the axis of the circle 3 cm distance from the centre.Correct answer is '1000'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle of mass 20 gm is moving in a circle of 4cm radius with contant speed of 10 cm/sec. What is the angular momentum (gm. cm2 / sec) about a point on the axis of the circle 3 cm distance from the centre.Correct answer is '1000'. Can you explain this answer?.
A particle of mass 20 gm is moving in a circle of 4cm radius with contant speed of 10 cm/sec. What is the angular momentum (gm. cm2 / sec) about a point on the axis of the circle 3 cm distance from the centre.Correct answer is '1000'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A particle of mass 20 gm is moving in a circle of 4cm radius with contant speed of 10 cm/sec. What is the angular momentum (gm. cm2 / sec) about a point on the axis of the circle 3 cm distance from the centre.Correct answer is '1000'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle of mass 20 gm is moving in a circle of 4cm radius with contant speed of 10 cm/sec. What is the angular momentum (gm. cm2 / sec) about a point on the axis of the circle 3 cm distance from the centre.Correct answer is '1000'. Can you explain this answer?.
Solutions for A particle of mass 20 gm is moving in a circle of 4cm radius with contant speed of 10 cm/sec. What is the angular momentum (gm. cm2 / sec) about a point on the axis of the circle 3 cm distance from the centre.Correct answer is '1000'. Can you explain this answer? in English & in Hindi are available as part of our courses for Physics.
Download more important topics, notes, lectures and mock test series for Physics Exam by signing up for free.
Here you can find the meaning of A particle of mass 20 gm is moving in a circle of 4cm radius with contant speed of 10 cm/sec. What is the angular momentum (gm. cm2 / sec) about a point on the axis of the circle 3 cm distance from the centre.Correct answer is '1000'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A particle of mass 20 gm is moving in a circle of 4cm radius with contant speed of 10 cm/sec. What is the angular momentum (gm. cm2 / sec) about a point on the axis of the circle 3 cm distance from the centre.Correct answer is '1000'. Can you explain this answer?, a detailed solution for A particle of mass 20 gm is moving in a circle of 4cm radius with contant speed of 10 cm/sec. What is the angular momentum (gm. cm2 / sec) about a point on the axis of the circle 3 cm distance from the centre.Correct answer is '1000'. Can you explain this answer? has been provided alongside types of A particle of mass 20 gm is moving in a circle of 4cm radius with contant speed of 10 cm/sec. What is the angular momentum (gm. cm2 / sec) about a point on the axis of the circle 3 cm distance from the centre.Correct answer is '1000'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A particle of mass 20 gm is moving in a circle of 4cm radius with contant speed of 10 cm/sec. What is the angular momentum (gm. cm2 / sec) about a point on the axis of the circle 3 cm distance from the centre.Correct answer is '1000'. Can you explain this answer? tests, examples and also practice Physics tests.
|
|
Explore Courses for Physics 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