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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?
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
2 ≥ μg
2 ≥  constant, or 
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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.
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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?
Question Description
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