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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 ms-2)
- a)2 kg- m2
- b)1 kg- m2
- c)4 kg- m2
- d)8 kg- m2
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
A wheel of radius 0.4 m can rotate freely about its axis as shown in t...
Option A is correct. Solution; radius=0.4 m
acceleration(alpha)=8 rads^-2 mass=4kg Torque=I×alpha. but torgue=m×g×r m×g×r=I× alpha 4×10×0.4=I×8 solve that u get I=2kg m^-2.
Community Answer
A wheel of radius 0.4 m can rotate freely about its axis as shown in t...
To find the moment of inertia of the wheel, we need to use the equation:
Torque (τ) = Moment of inertia (I) × Angular acceleration (α)
Given that the angular acceleration (α) is 8 rad/s^2 and the torque (τ) is the force (F) multiplied by the radius (r), we can rewrite the equation as:
F × r = I × α
We are given the radius of the wheel (r) as 0.4 m and the force (F) as the weight of the mass (m) hung on the string, which is 4 kg multiplied by the acceleration due to gravity (g = 10 m/s^2). Substituting these values into the equation, we have:
(4 kg × 10 m/s^2) × 0.4 m = I × 8 rad/s^2
Simplifying the equation:
(40 N) × 0.4 m = I × 8 rad/s^2
16 Nm = I × 8 rad/s^2
Dividing both sides of the equation by 8 rad/s^2:
I = 2 kg·m^2
Therefore, the moment of inertia of the wheel is 2 kg·m^2.
Torque (τ) = Moment of inertia (I) × Angular acceleration (α)
Given that the angular acceleration (α) is 8 rad/s^2 and the torque (τ) is the force (F) multiplied by the radius (r), we can rewrite the equation as:
F × r = I × α
We are given the radius of the wheel (r) as 0.4 m and the force (F) as the weight of the mass (m) hung on the string, which is 4 kg multiplied by the acceleration due to gravity (g = 10 m/s^2). Substituting these values into the equation, we have:
(4 kg × 10 m/s^2) × 0.4 m = I × 8 rad/s^2
Simplifying the equation:
(40 N) × 0.4 m = I × 8 rad/s^2
16 Nm = I × 8 rad/s^2
Dividing both sides of the equation by 8 rad/s^2:
I = 2 kg·m^2
Therefore, the moment of inertia of the wheel is 2 kg·m^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 is wrapped over its rim and a mass of 4 kg is hung. An angular acceleration of 8 rad- s-2is produced in it due to the torque. Then, moment of inertia of the wheel is (g = 10 ms-2)a)2 kg- m2b)1 kg- m2c)4 kg- m2d)8 kg- m2Correct answer is option 'A'. Can you explain this answer?
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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-2is produced in it due to the torque. Then, moment of inertia of the wheel is (g = 10 ms-2)a)2 kg- m2b)1 kg- m2c)4 kg- m2d)8 kg- m2Correct answer is option 'A'. 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 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-2is produced in it due to the torque. Then, moment of inertia of the wheel is (g = 10 ms-2)a)2 kg- m2b)1 kg- m2c)4 kg- m2d)8 kg- m2Correct answer is option 'A'. 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 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-2is produced in it due to the torque. Then, moment of inertia of the wheel is (g = 10 ms-2)a)2 kg- m2b)1 kg- m2c)4 kg- m2d)8 kg- m2Correct answer is option 'A'. Can you explain this answer?.
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