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A small object placed on a rotating horizontal turn table just slips when it is placed at a distance 4 cm from the axis of rotation. If the angular velocity of the turn-table is doubled, the object slips when its distance from the axis of rotation is
- a)1 cm
- b)2 cm
- c)4 cm
- d)8 cm
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A small object placed on a rotating horizontal turn table just slips w...
The object will slip if centripetal force ≥ force of friction
mrω2 ≥ μmg
rω2 ≥ μg
rω2 ≥ constant, or
mrω2 ≥ μmg
rω2 ≥ μg
rω2 ≥ constant, or
Most Upvoted Answer
A small object placed on a rotating horizontal turn table just slips w...
Understanding the Problem
A small object placed on a rotating turntable experiences centripetal force due to its rotation. The problem states that the object slips when it is 4 cm from the axis of rotation. When the angular velocity is doubled, we need to determine the new distance at which the object will slip.
Centripetal Force and Angular Velocity
- The centripetal force required to keep the object moving in a circle is given by the formula:
F_c = m * ω² * r
where:
- F_c is the centripetal force,
- m is the mass of the object,
- ω is the angular velocity, and
- r is the distance from the axis of rotation.
Slipping Condition
- The object will slip when the required centripetal force exceeds the maximum static friction force (F_friction).
- When the angular velocity is doubled (2ω), the new centripetal force becomes:
F_c' = m * (2ω)² * r = 4 * m * ω² * r.
New Distance Calculation
- To maintain equilibrium without slipping, the centripetal force must equal the frictional force. Assuming the maximum static friction remains constant, we can set up the equation:
F_friction = 4 * m * ω² * r'.
- Given that the object slips at 4 cm when ω is normal, we can express the relationship:
m * ω² * 4 cm = F_friction.
- When the angular velocity is doubled, we want to find r' such that:
F_friction = 4 * m * ω² * r'.
- Equating both expressions for F_friction results in:
4 * m * ω² * r' = m * ω² * 4 cm, which simplifies to:
r' = 1 cm.
Conclusion
- Thus, when the angular velocity of the turntable is doubled, the object will slip at a distance of 1 cm from the axis of rotation, confirming that the correct answer is option A.
A small object placed on a rotating turntable experiences centripetal force due to its rotation. The problem states that the object slips when it is 4 cm from the axis of rotation. When the angular velocity is doubled, we need to determine the new distance at which the object will slip.
Centripetal Force and Angular Velocity
- The centripetal force required to keep the object moving in a circle is given by the formula:
F_c = m * ω² * r
where:
- F_c is the centripetal force,
- m is the mass of the object,
- ω is the angular velocity, and
- r is the distance from the axis of rotation.
Slipping Condition
- The object will slip when the required centripetal force exceeds the maximum static friction force (F_friction).
- When the angular velocity is doubled (2ω), the new centripetal force becomes:
F_c' = m * (2ω)² * r = 4 * m * ω² * r.
New Distance Calculation
- To maintain equilibrium without slipping, the centripetal force must equal the frictional force. Assuming the maximum static friction remains constant, we can set up the equation:
F_friction = 4 * m * ω² * r'.
- Given that the object slips at 4 cm when ω is normal, we can express the relationship:
m * ω² * 4 cm = F_friction.
- When the angular velocity is doubled, we want to find r' such that:
F_friction = 4 * m * ω² * r'.
- Equating both expressions for F_friction results in:
4 * m * ω² * r' = m * ω² * 4 cm, which simplifies to:
r' = 1 cm.
Conclusion
- Thus, when the angular velocity of the turntable is doubled, the object will slip at a distance of 1 cm from the axis of rotation, confirming that the correct answer is option A.
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A small object placed on a rotating horizontal turn table just slips when it is placed at a distance 4 cm from the axis of rotation. If the angular velocity of the turn-table is doubled, the object slips when its distance from the axis of rotation isa)1 cmb)2 cmc)4 cmd)8 cmCorrect answer is option 'A'. Can you explain this answer?
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A small object placed on a rotating horizontal turn table just slips when it is placed at a distance 4 cm from the axis of rotation. If the angular velocity of the turn-table is doubled, the object slips when its distance from the axis of rotation isa)1 cmb)2 cmc)4 cmd)8 cmCorrect 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 small object placed on a rotating horizontal turn table just slips when it is placed at a distance 4 cm from the axis of rotation. If the angular velocity of the turn-table is doubled, the object slips when its distance from the axis of rotation isa)1 cmb)2 cmc)4 cmd)8 cmCorrect 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 small object placed on a rotating horizontal turn table just slips when it is placed at a distance 4 cm from the axis of rotation. If the angular velocity of the turn-table is doubled, the object slips when its distance from the axis of rotation isa)1 cmb)2 cmc)4 cmd)8 cmCorrect answer is option 'A'. Can you explain this answer?.
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