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The period of a disk of radius 10.2 cm executing small oscillation about a pivot at its rim is measured to be _____ sec. { The value o f the all acceleration due to gravity at that location is 9.83 m/sec2}
Correct answer is '0.784'. Can you explain this answer?
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Verified Answer
The period of a disk of radius 10.2 cm executing small oscillation abo...
0.6-0.8
The rotational inertia of a disk about an axis through its centre is
where R is the radius and M is the mass of the disk. The rotational inertia about the pivot at the rim is, using the parallel axis theorem
The period of this physical pendulum T =
With d = R, is then
independent of the mass of the disk The simple pendulum having the same period has a length
The rotational inertia of a disk about an axis through its centre is
where R is the radius and M is the mass of the disk. The rotational inertia about the pivot at the rim is, using the parallel axis theoremThe period of this physical pendulum T =
With d = R, is then
independent of the mass of the disk The simple pendulum having the same period has a length
Most Upvoted Answer
The period of a disk of radius 10.2 cm executing small oscillation abo...
Given:
- Radius of the disk: 10.2 cm
- Acceleration due to gravity: 9.83 m/s^2
To find:
- Period of oscillation of the disk
Formula:
The period of oscillation for a simple pendulum is given by the formula:
T = 2π √(L/g)
where:
T = Period of oscillation
L = Length of the pendulum
g = Acceleration due to gravity
Conversion:
To use the formula, we need to convert the radius of the disk to length.
Length = 2πr
Calculation:
Given that the radius of the disk is 10.2 cm, we can calculate the length as follows:
Length = 2π(10.2 cm) = 64.08 cm
Now, we need to convert the length to meters:
Length = 64.08 cm × (1 m/100 cm) = 0.6408 m
Substituting the values into the formula, we can calculate the period of oscillation:
T = 2π √(0.6408 m/9.83 m/s^2)
T = 2π √(0.0653)
T ≈ 2π(0.2554)
T ≈ 1.605 s
Therefore, the period of oscillation of the disk is approximately 1.605 seconds.
Answer:
The period of a disk of radius 10.2 cm executing small oscillation about a pivot at its rim is approximately 1.605 seconds.
- Radius of the disk: 10.2 cm
- Acceleration due to gravity: 9.83 m/s^2
To find:
- Period of oscillation of the disk
Formula:
The period of oscillation for a simple pendulum is given by the formula:
T = 2π √(L/g)
where:
T = Period of oscillation
L = Length of the pendulum
g = Acceleration due to gravity
Conversion:
To use the formula, we need to convert the radius of the disk to length.
Length = 2πr
Calculation:
Given that the radius of the disk is 10.2 cm, we can calculate the length as follows:
Length = 2π(10.2 cm) = 64.08 cm
Now, we need to convert the length to meters:
Length = 64.08 cm × (1 m/100 cm) = 0.6408 m
Substituting the values into the formula, we can calculate the period of oscillation:
T = 2π √(0.6408 m/9.83 m/s^2)
T = 2π √(0.0653)
T ≈ 2π(0.2554)
T ≈ 1.605 s
Therefore, the period of oscillation of the disk is approximately 1.605 seconds.
Answer:
The period of a disk of radius 10.2 cm executing small oscillation about a pivot at its rim is approximately 1.605 seconds.
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The period of a disk of radius 10.2 cm executing small oscillation about a pivot at its rim is measured to be _____ sec. { The value o f the all acceleration due to gravity at that location is 9.83 m/sec2}Correct answer is '0.784'. Can you explain this answer?
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The period of a disk of radius 10.2 cm executing small oscillation about a pivot at its rim is measured to be _____ sec. { The value o f the all acceleration due to gravity at that location is 9.83 m/sec2}Correct answer is '0.784'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about The period of a disk of radius 10.2 cm executing small oscillation about a pivot at its rim is measured to be _____ sec. { The value o f the all acceleration due to gravity at that location is 9.83 m/sec2}Correct answer is '0.784'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The period of a disk of radius 10.2 cm executing small oscillation about a pivot at its rim is measured to be _____ sec. { The value o f the all acceleration due to gravity at that location is 9.83 m/sec2}Correct answer is '0.784'. Can you explain this answer?.
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