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A wheel of radius 0.4 m can rotate freely about its axis as shown in the figure. A string iswrapped over its rim and a mass of 4 kg is hung. An angular acceleration of 8 rad- s-2 is produced init due to the torque. Then, moment of inertia of the wheel is (g = 10 ms-2)
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A wheel of radius 0.4 m can rotate freely about its axis as shown in t...
From FBD of block Mg-Tension=ma40-T=4a40-T=4Xr where X is angular acceleration r is radius( a=Xr) After substituting the value we get 40-T=4x8x. 440 - T=12.8T=27.2NNow torque ¬=TxrAnd torque ¬=I.X .... (I=inertia) Substitute the values in Txr=I. X27.2x .4=I x 8Then I=13.6 Nm ^2According to me answer should be this if there is any mistake Plz notify me..
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A wheel of radius 0.4 m can rotate freely about its axis as shown in t...
Problem:
A wheel of radius 0.4 m can rotate freely about its axis as shown in the figure. A string is wrapped over its rim and a mass of 4 kg is hung. An angular acceleration of 8 rad/s^2 is produced in it due to the torque. Then, moment of inertia of the wheel is (g = 10 m/s^2).
Solution:
Given data:
- Radius of the wheel (r) = 0.4 m
- Mass of the hanging object (m) = 4 kg
- Angular acceleration (α) = 8 rad/s^2
- Acceleration due to gravity (g) = 10 m/s^2
Analysis:
To find the moment of inertia of the wheel, we need to use the formula:
Torque (τ) = Moment of inertia (I) * Angular acceleration (α)
The torque acting on the wheel is given by:
τ = r * F
where F is the force acting on the wheel due to the hanging mass.
We can calculate the force F using Newton's second law:
F = m * g
Calculations:
1. Calculate the force acting on the wheel:
F = m * g
= 4 kg * 10 m/s^2
= 40 N
2. Calculate the torque acting on the wheel:
τ = r * F
= 0.4 m * 40 N
= 16 Nm
3. Use the formula to find the moment of inertia:
τ = I * α
16 Nm = I * 8 rad/s^2
4. Rearrange the equation to solve for I:
I = τ / α
I = 16 Nm / 8 rad/s^2
I = 2 kgm^2
Answer:
The moment of inertia of the wheel is 2 kgm^2.
A wheel of radius 0.4 m can rotate freely about its axis as shown in the figure. A string is wrapped over its rim and a mass of 4 kg is hung. An angular acceleration of 8 rad/s^2 is produced in it due to the torque. Then, moment of inertia of the wheel is (g = 10 m/s^2).
Solution:
Given data:
- Radius of the wheel (r) = 0.4 m
- Mass of the hanging object (m) = 4 kg
- Angular acceleration (α) = 8 rad/s^2
- Acceleration due to gravity (g) = 10 m/s^2
Analysis:
To find the moment of inertia of the wheel, we need to use the formula:
Torque (τ) = Moment of inertia (I) * Angular acceleration (α)
The torque acting on the wheel is given by:
τ = r * F
where F is the force acting on the wheel due to the hanging mass.
We can calculate the force F using Newton's second law:
F = m * g
Calculations:
1. Calculate the force acting on the wheel:
F = m * g
= 4 kg * 10 m/s^2
= 40 N
2. Calculate the torque acting on the wheel:
τ = r * F
= 0.4 m * 40 N
= 16 Nm
3. Use the formula to find the moment of inertia:
τ = I * α
16 Nm = I * 8 rad/s^2
4. Rearrange the equation to solve for I:
I = τ / α
I = 16 Nm / 8 rad/s^2
I = 2 kgm^2
Answer:
The moment of inertia of the wheel is 2 kgm^2.
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A wheel of radius 0.4 m can rotate freely about its axis as shown in the figure. A string iswrapped over its rim and a mass of 4 kg is hung. An angular acceleration of 8 rad- s-2 is produced init due to the torque. Then, moment of inertia of the wheel is (g = 10 ms-2)
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A wheel of radius 0.4 m can rotate freely about its axis as shown in the figure. A string iswrapped over its rim and a mass of 4 kg is hung. An angular acceleration of 8 rad- s-2 is produced init due to the torque. Then, moment of inertia of the wheel is (g = 10 ms-2) for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A wheel of radius 0.4 m can rotate freely about its axis as shown in the figure. A string iswrapped over its rim and a mass of 4 kg is hung. An angular acceleration of 8 rad- s-2 is produced init due to the torque. Then, moment of inertia of the wheel is (g = 10 ms-2) covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A wheel of radius 0.4 m can rotate freely about its axis as shown in the figure. A string iswrapped over its rim and a mass of 4 kg is hung. An angular acceleration of 8 rad- s-2 is produced init due to the torque. Then, moment of inertia of the wheel is (g = 10 ms-2).
A wheel of radius 0.4 m can rotate freely about its axis as shown in the figure. A string iswrapped over its rim and a mass of 4 kg is hung. An angular acceleration of 8 rad- s-2 is produced init due to the torque. Then, moment of inertia of the wheel is (g = 10 ms-2) for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A wheel of radius 0.4 m can rotate freely about its axis as shown in the figure. A string iswrapped over its rim and a mass of 4 kg is hung. An angular acceleration of 8 rad- s-2 is produced init due to the torque. Then, moment of inertia of the wheel is (g = 10 ms-2) covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A wheel of radius 0.4 m can rotate freely about its axis as shown in the figure. A string iswrapped over its rim and a mass of 4 kg is hung. An angular acceleration of 8 rad- s-2 is produced init due to the torque. Then, moment of inertia of the wheel is (g = 10 ms-2).
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A wheel of radius 0.4 m can rotate freely about its axis as shown in the figure. A string iswrapped over its rim and a mass of 4 kg is hung. An angular acceleration of 8 rad- s-2 is produced init due to the torque. Then, moment of inertia of the wheel is (g = 10 ms-2), a detailed solution for A wheel of radius 0.4 m can rotate freely about its axis as shown in the figure. A string iswrapped over its rim and a mass of 4 kg is hung. An angular acceleration of 8 rad- s-2 is produced init due to the torque. Then, moment of inertia of the wheel is (g = 10 ms-2) has been provided alongside types of A wheel of radius 0.4 m can rotate freely about its axis as shown in the figure. A string iswrapped over its rim and a mass of 4 kg is hung. An angular acceleration of 8 rad- s-2 is produced init due to the torque. Then, moment of inertia of the wheel is (g = 10 ms-2) theory, EduRev gives you an
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