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A 25 kg uniform solid sphere with a 20 cm radius is suspended by a vertical wire such that the point of suspension is vertically above the centre of the sphere. A torque of 0.10 N-m is required to rotate the sphere through an angle of 1.0 rad and then maintain the orientation. If the sphere is then released, its time period of the oscillation will be:
- a)(π sec)
- b)(4π sec)
- c)(2π sec)
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
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Assuming small angle oscillations, the time period of the oscillation can be calculated using the formula:
T = 2π√(I/mgd)
where I is the moment of inertia of the sphere, m is its mass, g is the acceleration due to gravity, and d is the distance between the point of suspension and the centre of mass of the sphere.
The moment of inertia of a solid sphere is given by:
I = (2/5)mr^2
where r is the radius of the sphere.
Substituting the given values, we get:
I = (2/5)(25 kg)(0.20 m)^2 = 0.50 kg·m^2
m = 25 kg
g = 9.81 m/s^2
d = 0.20 m
To find the angular frequency, we need to first find the torque constant k:
k = τ/θ = 0.10 N-m/1.0 rad = 0.10 N-m/rad
The torque constant is related to the moment of inertia and the angular frequency (ω) by:
k = Iω^2
Substituting the values, we get:
0.10 N-m/rad = (0.50 kg·m^2)ω^2
Solving for ω, we get:
ω = √(0.20 N/m) = 0.45 rad/s
Finally, substituting the values in the formula for time period, we get:
T = 2π√(I/mgd) = 2π√[(0.50 kg·m^2)/(25 kg)(9.81 m/s^2)(0.20 m)] = 1.33 s
Therefore, the time period of the oscillation is 1.33 s.
T = 2π√(I/mgd)
where I is the moment of inertia of the sphere, m is its mass, g is the acceleration due to gravity, and d is the distance between the point of suspension and the centre of mass of the sphere.
The moment of inertia of a solid sphere is given by:
I = (2/5)mr^2
where r is the radius of the sphere.
Substituting the given values, we get:
I = (2/5)(25 kg)(0.20 m)^2 = 0.50 kg·m^2
m = 25 kg
g = 9.81 m/s^2
d = 0.20 m
To find the angular frequency, we need to first find the torque constant k:
k = τ/θ = 0.10 N-m/1.0 rad = 0.10 N-m/rad
The torque constant is related to the moment of inertia and the angular frequency (ω) by:
k = Iω^2
Substituting the values, we get:
0.10 N-m/rad = (0.50 kg·m^2)ω^2
Solving for ω, we get:
ω = √(0.20 N/m) = 0.45 rad/s
Finally, substituting the values in the formula for time period, we get:
T = 2π√(I/mgd) = 2π√[(0.50 kg·m^2)/(25 kg)(9.81 m/s^2)(0.20 m)] = 1.33 s
Therefore, the time period of the oscillation is 1.33 s.
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A 25 kg uniform solid sphere with a 20 cm radius is suspended by a vertical wire such that the point of suspension is vertically above the centre of the sphere. A torque of 0.10 N-m is required to rotate the sphere through an angle of 1.0 rad and then maintain the orientation. If the sphere is then released, its time period of the oscillation will be:a)(π sec)b)(4π sec)c)(2π sec)d)None of theseCorrect answer is option 'B'. Can you explain this answer?
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A 25 kg uniform solid sphere with a 20 cm radius is suspended by a vertical wire such that the point of suspension is vertically above the centre of the sphere. A torque of 0.10 N-m is required to rotate the sphere through an angle of 1.0 rad and then maintain the orientation. If the sphere is then released, its time period of the oscillation will be:a)(π sec)b)(4π sec)c)(2π sec)d)None of theseCorrect answer is option 'B'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A 25 kg uniform solid sphere with a 20 cm radius is suspended by a vertical wire such that the point of suspension is vertically above the centre of the sphere. A torque of 0.10 N-m is required to rotate the sphere through an angle of 1.0 rad and then maintain the orientation. If the sphere is then released, its time period of the oscillation will be:a)(π sec)b)(4π sec)c)(2π sec)d)None of theseCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 25 kg uniform solid sphere with a 20 cm radius is suspended by a vertical wire such that the point of suspension is vertically above the centre of the sphere. A torque of 0.10 N-m is required to rotate the sphere through an angle of 1.0 rad and then maintain the orientation. If the sphere is then released, its time period of the oscillation will be:a)(π sec)b)(4π sec)c)(2π sec)d)None of theseCorrect answer is option 'B'. Can you explain this answer?.
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