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A uniform disk of mass m and radius R rolls, without slipping, down a fixed plane inclined at an angle 30° to the horizontal. The linear acceleration of the disk (in m/sec2) is _______________.
Correct answer is '3.266'. Can you explain this answer?
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A uniform disk of mass m and radius R rolls, without slipping, down a ...
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Solution:
Acceleration of the rolling body can be calculated by considering the rolling motion and linear motion separately.
Rolling Motion:
- For pure rolling, the velocity of the center of mass of the disk is equal to the velocity of the point of contact with the plane, i.e., v = Rω.
- The angular velocity of the disk can be calculated using the conservation of energy principle. The initial potential energy is mgh, and the final energy is a combination of kinetic energy of the center of mass and rotational energy about the center of mass. So, mgh = (1/2)mv^2 + (1/2)Iω^2, where I is the moment of inertia of the disk, which is (1/2)mr^2.
- Substituting v = Rω and I = (1/2)mr^2 in the above equation and solving for ω, we get ω = (2gh/R)^0.5.
Linear Motion:
- The gravitational force acting on the disk is mg sin(30), where 30 is the angle of inclination of the plane.
- The frictional force acting on the disk is mg cos(30) μ, where μ is the coefficient of static friction between the disk and the plane.
- The net force acting on the disk is F = mg sin(30) - mg cos(30) μ.
- Using Newton's second law, F = ma, where a is the linear acceleration of the disk.
- Substituting the value of F and solving for a, we get a = g (sin(30) - cos(30) μ).
Final Answer:
- Substituting the value of μ = 0.5 and g = 9.8 m/s^2 in the above equation, we get a = 3.266 m/s^2 (approx).
Acceleration of the rolling body can be calculated by considering the rolling motion and linear motion separately.
Rolling Motion:
- For pure rolling, the velocity of the center of mass of the disk is equal to the velocity of the point of contact with the plane, i.e., v = Rω.
- The angular velocity of the disk can be calculated using the conservation of energy principle. The initial potential energy is mgh, and the final energy is a combination of kinetic energy of the center of mass and rotational energy about the center of mass. So, mgh = (1/2)mv^2 + (1/2)Iω^2, where I is the moment of inertia of the disk, which is (1/2)mr^2.
- Substituting v = Rω and I = (1/2)mr^2 in the above equation and solving for ω, we get ω = (2gh/R)^0.5.
Linear Motion:
- The gravitational force acting on the disk is mg sin(30), where 30 is the angle of inclination of the plane.
- The frictional force acting on the disk is mg cos(30) μ, where μ is the coefficient of static friction between the disk and the plane.
- The net force acting on the disk is F = mg sin(30) - mg cos(30) μ.
- Using Newton's second law, F = ma, where a is the linear acceleration of the disk.
- Substituting the value of F and solving for a, we get a = g (sin(30) - cos(30) μ).
Final Answer:
- Substituting the value of μ = 0.5 and g = 9.8 m/s^2 in the above equation, we get a = 3.266 m/s^2 (approx).
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A uniform disk of mass m and radius R rolls, without slipping, down a fixed plane inclined at an angle 30 to the horizontal. The linear acceleration of the disk (in m/sec2) is _______________.Correct answer is '3.266'. Can you explain this answer?
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A uniform disk of mass m and radius R rolls, without slipping, down a fixed plane inclined at an angle 30 to the horizontal. The linear acceleration of the disk (in m/sec2) is _______________.Correct answer is '3.266'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about A uniform disk of mass m and radius R rolls, without slipping, down a fixed plane inclined at an angle 30 to the horizontal. The linear acceleration of the disk (in m/sec2) is _______________.Correct answer is '3.266'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A uniform disk of mass m and radius R rolls, without slipping, down a fixed plane inclined at an angle 30 to the horizontal. The linear acceleration of the disk (in m/sec2) is _______________.Correct answer is '3.266'. Can you explain this answer?.
A uniform disk of mass m and radius R rolls, without slipping, down a fixed plane inclined at an angle 30 to the horizontal. The linear acceleration of the disk (in m/sec2) is _______________.Correct answer is '3.266'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about A uniform disk of mass m and radius R rolls, without slipping, down a fixed plane inclined at an angle 30 to the horizontal. The linear acceleration of the disk (in m/sec2) is _______________.Correct answer is '3.266'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A uniform disk of mass m and radius R rolls, without slipping, down a fixed plane inclined at an angle 30 to the horizontal. The linear acceleration of the disk (in m/sec2) is _______________.Correct answer is '3.266'. Can you explain this answer?.
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