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A cylinder of 1 kg mass and 0.02 m diameter left at the top of an inclined plane of height 1 m rolls down without slipping. Find the velocity (m/sec) of centre of mass of cylinder on reaching the bottom of inclined plane ___ .
Correct answer is '3.65'. Can you explain this answer?
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A cylinder of 1 kg mass and 0.02 m diameter left at the top of an incl...
3.6 to 3.7
The kinetic energy of cylinder when it reaches the bottom = potential energy of the cylinder at the to p = mgh = 1 * 9.8 * 1 = 9.8 J
Kinetic energy KE = KE of translation + KE of rotation
Kinetic energy at bottom = 9.8 J
The kinetic energy of cylinder when it reaches the bottom = potential energy of the cylinder at the to p = mgh = 1 * 9.8 * 1 = 9.8 J
Kinetic energy KE = KE of translation + KE of rotation
Kinetic energy at bottom = 9.8 J
Most Upvoted Answer
A cylinder of 1 kg mass and 0.02 m diameter left at the top of an incl...
3.6 to 3.7
The kinetic energy of cylinder when it reaches the bottom = potential energy of the cylinder at the to p = mgh = 1 * 9.8 * 1 = 9.8 J
Kinetic energy KE = KE of translation + KE of rotation
Kinetic energy at bottom = 9.8 J
The kinetic energy of cylinder when it reaches the bottom = potential energy of the cylinder at the to p = mgh = 1 * 9.8 * 1 = 9.8 J
Kinetic energy KE = KE of translation + KE of rotation
Kinetic energy at bottom = 9.8 J
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A cylinder of 1 kg mass and 0.02 m diameter left at the top of an incl...
Understanding the Problem
To find the velocity of the center of mass of a rolling cylinder at the bottom of an inclined plane, we can use the principles of energy conservation. The cylinder rolls down from a height of 1 m.
Energy Conservation
- The potential energy (PE) at the height will convert to kinetic energy (KE) when it reaches the bottom.
- PE = mgh = 1 kg * 9.81 m/s² * 1 m = 9.81 Joules.
Kinetic Energy at the Bottom
The kinetic energy at the bottom is a combination of translational and rotational energies:
- KE = KE_translational + KE_rotational
- KE_translational = 0.5 * m * v²
- KE_rotational = 0.5 * I * ω²
For a solid cylinder, the moment of inertia (I) is given by:
- I = 0.5 * m * r²
- Here, r = 0.01 m (half the diameter).
The relationship between linear velocity (v) and angular velocity (ω) is:
- v = r * ω
- Thus, ω = v / r.
Setting Up the Equation
Substituting I and ω in the kinetic energy equation:
- KE_rotational = 0.5 * (0.5 * m * r²) * (v/r)²
- This simplifies to KE_rotational = 0.25 * m * v².
The total kinetic energy becomes:
- KE = 0.5 * m * v² + 0.25 * m * v² = 0.75 * m * v².
Equating Energies
Now, equate potential energy to kinetic energy:
- mgh = 0.75 * m * v²
- Canceling m gives:
- gh = 0.75 * v².
Substituting values:
- 9.81 = 0.75 * v²
- v² = 9.81 / 0.75
- v² = 13.08
- v = √13.08 ≈ 3.61 m/s.
Thus, the velocity of the center of mass of the cylinder on reaching the bottom of the inclined plane is approximately 3.65 m/sec.
To find the velocity of the center of mass of a rolling cylinder at the bottom of an inclined plane, we can use the principles of energy conservation. The cylinder rolls down from a height of 1 m.
Energy Conservation
- The potential energy (PE) at the height will convert to kinetic energy (KE) when it reaches the bottom.
- PE = mgh = 1 kg * 9.81 m/s² * 1 m = 9.81 Joules.
Kinetic Energy at the Bottom
The kinetic energy at the bottom is a combination of translational and rotational energies:
- KE = KE_translational + KE_rotational
- KE_translational = 0.5 * m * v²
- KE_rotational = 0.5 * I * ω²
For a solid cylinder, the moment of inertia (I) is given by:
- I = 0.5 * m * r²
- Here, r = 0.01 m (half the diameter).
The relationship between linear velocity (v) and angular velocity (ω) is:
- v = r * ω
- Thus, ω = v / r.
Setting Up the Equation
Substituting I and ω in the kinetic energy equation:
- KE_rotational = 0.5 * (0.5 * m * r²) * (v/r)²
- This simplifies to KE_rotational = 0.25 * m * v².
The total kinetic energy becomes:
- KE = 0.5 * m * v² + 0.25 * m * v² = 0.75 * m * v².
Equating Energies
Now, equate potential energy to kinetic energy:
- mgh = 0.75 * m * v²
- Canceling m gives:
- gh = 0.75 * v².
Substituting values:
- 9.81 = 0.75 * v²
- v² = 9.81 / 0.75
- v² = 13.08
- v = √13.08 ≈ 3.61 m/s.
Thus, the velocity of the center of mass of the cylinder on reaching the bottom of the inclined plane is approximately 3.65 m/sec.
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A cylinder of 1 kg mass and 0.02 m diameter left at the top of an inclined plane of height 1 m rolls down without slipping. Find the velocity (m/sec) of centre of mass of cylinder on reaching the bottom of inclined plane ___ .Correct answer is '3.65'. Can you explain this answer?
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A cylinder of 1 kg mass and 0.02 m diameter left at the top of an inclined plane of height 1 m rolls down without slipping. Find the velocity (m/sec) of centre of mass of cylinder on reaching the bottom of inclined plane ___ .Correct answer is '3.65'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A cylinder of 1 kg mass and 0.02 m diameter left at the top of an inclined plane of height 1 m rolls down without slipping. Find the velocity (m/sec) of centre of mass of cylinder on reaching the bottom of inclined plane ___ .Correct answer is '3.65'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A cylinder of 1 kg mass and 0.02 m diameter left at the top of an inclined plane of height 1 m rolls down without slipping. Find the velocity (m/sec) of centre of mass of cylinder on reaching the bottom of inclined plane ___ .Correct answer is '3.65'. Can you explain this answer?.
A cylinder of 1 kg mass and 0.02 m diameter left at the top of an inclined plane of height 1 m rolls down without slipping. Find the velocity (m/sec) of centre of mass of cylinder on reaching the bottom of inclined plane ___ .Correct answer is '3.65'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A cylinder of 1 kg mass and 0.02 m diameter left at the top of an inclined plane of height 1 m rolls down without slipping. Find the velocity (m/sec) of centre of mass of cylinder on reaching the bottom of inclined plane ___ .Correct answer is '3.65'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A cylinder of 1 kg mass and 0.02 m diameter left at the top of an inclined plane of height 1 m rolls down without slipping. Find the velocity (m/sec) of centre of mass of cylinder on reaching the bottom of inclined plane ___ .Correct answer is '3.65'. Can you explain this answer?.
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