Take Class 11 Tuition from the Best Tutors
Search in
Answered on 14 Apr Learn Unit 10-Oscillation & Waves
Nazia Khanum
As an experienced tutor registered on UrbanPro, I'd be glad to assist you with this physics problem.
Given that a circular disc of mass 10 kg is suspended by a wire attached to its center, and the wire is twisted by rotating the disc and released, resulting in torsional oscillations with a period of 1.5 seconds. The radius of the disc is 15 cm.
To determine the torsional spring constant (α) of the wire, we can use the equation provided:
J=−αθJ=−αθ
Where:
The restoring couple JJ can be related to the torque acting on the disc, which is given by:
J=I⋅αJ=I⋅α
Where:
The moment of inertia of a circular disc about its center is given by:
I=12mr2I=21mr2
Where:
Given that m=10m=10 kg and r=15r=15 cm, we can calculate the moment of inertia II.
I=12×10×(0.15)2I=21×10×(0.15)2 I=0.1125 kg m2I=0.1125 kg m2
Now, substituting II into the equation for the restoring couple:
J=0.1125×αJ=0.1125×α
Since the period of torsional oscillations (TT) is related to the angular frequency (ωω) by T=2πωT=ω2π, we can find ωω:
T=2πωT=ω2π ω=2πTω=T2π ω=2π1.5ω=1.52π ω≈4.19 rad/sω≈4.19 rad/s
Now, the relation between angular frequency (ωω) and torsional spring constant (αα) is:
ω=αIω=Iα
Substituting the known values:
4.19=α0.11254.19=0.1125α
Solving for αα:
α=(4.19)2×0.1125α=(4.19)2×0.1125 α≈1.86 Nm/radα≈1.86 Nm/rad
So, the torsional spring constant of the wire is approximately 1.86 Nm/rad1.86 Nm/rad.
Feel free to ask if you have any questions or need further clarification! And remember, UrbanPro is a great platform for finding excellent tutors for your academic needs.
Answered on 28 Apr Learn Unit 10-Oscillation & Waves
Deepika Agrawal
"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."
Answered on 28 Apr Learn Chapter 4-Motion in a Plane
Deepika Agrawal
"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."
Take Class 11 Tuition from the Best Tutors
Answered on 28 Apr Learn Unit 3-Laws of Motion
Deepika Agrawal
"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."
Answered on 28 Apr Learn Unit 5-Work, Energy and Power
Deepika Agrawal
"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."
Answered on 28 Apr Learn Unit 4-Motion of System of Particles
Deepika Agrawal
"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."
Take Class 11 Tuition from the Best Tutors
Answered on 28 Apr Learn Chapter 9-Mechanical Properties of Solids
Deepika Agrawal
"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."
Mass of the big structure, M = 50,000 kg
Inner radius of the column, r = 30 cm = 0.3 m
Outer radius of the column, R = 60 cm = 0.6 m
Young’s modulus of steel, Y=2×1011Pa
Total force exerted, F=Mg=50000×9.8N
Stress = Force exerted on a single column =50000×9.84=122500N
Young’s modulus, Y=StressStrain
Strain =(F/A)Y
Where,
Area, A=π(R2−r2)=π((0.6)2−(0.3)2)
Strain =122500/[π((0.6)2−(0.3)2)×2×1011]=7.22×10−7
Hence, the compressional strain of each column is 7.22×10−7.
read lessAnswered on 28 Apr Learn Unit 8-Thermodynamics
Deepika Agrawal
"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."
Answered on 28 Apr Learn Chapter 3-Motion in a Straight Line
Deepika Agrawal
"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."
Take Class 11 Tuition from the Best Tutors
Answered on 28 Apr Learn Chapter 3-Motion in a Straight Line
Deepika Agrawal
"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."
UrbanPro.com helps you to connect with the best Class 11 Tuition in India. Post Your Requirement today and get connected.
Ask a Question
The best tutors for Class 11 Tuition Classes are on UrbanPro
The best Tutors for Class 11 Tuition Classes are on UrbanPro