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A coin placed on a rotating turn table just slips if it is at a distance of 40 cm from the centre if the angular velocity of the turntable is doubled, it will just slip at a distance of
- a)10 cm
- b)20 cm
- c)40 cm
- d)80 cm
Correct answer is option 'A'. Can you explain this answer?
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Given:
- Distance of the coin from the center of the turntable = 40 cm
To find:
The distance at which the coin will just slip when the angular velocity of the turntable is doubled.
Solution:
When an object is placed on a rotating turntable, it experiences two types of forces:
1. Centripetal force (Fc): This force acts towards the center of the turntable and keeps the object moving in a circular path.
2. Frictional force (Ff): This force acts tangentially to the circular path and prevents the object from slipping off the turntable.
When the object is just at the point of slipping, the frictional force is equal to the maximum value it can attain, which is given by the equation:
Ff = μN
Where:
- Ff is the frictional force
- μ is the coefficient of friction
- N is the normal force
Since the coin is just at the point of slipping, the frictional force is equal to the maximum value it can attain. Therefore, we can equate the centripetal force to the frictional force:
Fc = Ff
Using the equation for centripetal force:
Fc = mv²/r
Where:
- Fc is the centripetal force
- m is the mass of the coin
- v is the linear velocity of the coin
- r is the distance of the coin from the center of the turntable
Substituting the values:
mv²/r = μN
Since the coin is in contact with the turntable, the normal force (N) is equal to the weight of the coin (mg), where g is the acceleration due to gravity.
mv²/r = μmg
Canceling out the mass (m) from both sides:
v²/r = μg
Since the angular velocity (ω) is given by:
ω = v/r
Substituting the value of v/r in terms of ω:
(ωr)²/r = μg
ω²r = μg
When the angular velocity is doubled:
(2ω)²r = μg
4ω²r = μg
Comparing the two equations:
ω²r = 4ω²r
This means that when the angular velocity is doubled, the distance at which the coin will just slip remains the same. Therefore, the correct answer is option A) 10 cm.
- Distance of the coin from the center of the turntable = 40 cm
To find:
The distance at which the coin will just slip when the angular velocity of the turntable is doubled.
Solution:
When an object is placed on a rotating turntable, it experiences two types of forces:
1. Centripetal force (Fc): This force acts towards the center of the turntable and keeps the object moving in a circular path.
2. Frictional force (Ff): This force acts tangentially to the circular path and prevents the object from slipping off the turntable.
When the object is just at the point of slipping, the frictional force is equal to the maximum value it can attain, which is given by the equation:
Ff = μN
Where:
- Ff is the frictional force
- μ is the coefficient of friction
- N is the normal force
Since the coin is just at the point of slipping, the frictional force is equal to the maximum value it can attain. Therefore, we can equate the centripetal force to the frictional force:
Fc = Ff
Using the equation for centripetal force:
Fc = mv²/r
Where:
- Fc is the centripetal force
- m is the mass of the coin
- v is the linear velocity of the coin
- r is the distance of the coin from the center of the turntable
Substituting the values:
mv²/r = μN
Since the coin is in contact with the turntable, the normal force (N) is equal to the weight of the coin (mg), where g is the acceleration due to gravity.
mv²/r = μmg
Canceling out the mass (m) from both sides:
v²/r = μg
Since the angular velocity (ω) is given by:
ω = v/r
Substituting the value of v/r in terms of ω:
(ωr)²/r = μg
ω²r = μg
When the angular velocity is doubled:
(2ω)²r = μg
4ω²r = μg
Comparing the two equations:
ω²r = 4ω²r
This means that when the angular velocity is doubled, the distance at which the coin will just slip remains the same. Therefore, the correct answer is option A) 10 cm.
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A coin placed on a rotating turn table just slips if it is at a distance of 40 cm from the centre if the angular velocity of the turntable is doubled, it will just slip at a distance ofa)10 cmb)20 cmc)40 cmd)80 cmCorrect answer is option 'A'. Can you explain this answer?
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