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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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