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Consider a mass m attached at radius r from the axis of shaft which rotates with an angular velocity ω. The balancing is achieved by mounting a B kg mass at radius bfrom the axis of shaft. If the speed of shaft is doubled, then to attain perfect balance, the value of mass B should be
- a)halved
- b)doubled
- c)quadrupled
- d)remain unchanged
Correct answer is option 'D'. Can you explain this answer?
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Consider a mass m attached at radius r from the axis of shaft which ro...
∴ The value of balancing mass does not depend upon speed of rotation.
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Consider a mass m attached at radius r from the axis of shaft which ro...
Ω. The centripetal force acting on the mass is given by F = mω^2r. This force is directed towards the center of rotation and provides the necessary inward force to keep the mass moving in a circular path.
The tension in the shaft is equal in magnitude and opposite in direction to the centripetal force. Therefore, the tension in the shaft is T = -mω^2r.
In order to find the maximum angular velocity ω_max at which the mass can be attached without breaking the shaft, we need to consider the maximum tension that the shaft can withstand.
Let's assume that the maximum tension the shaft can withstand is T_max. This means that the magnitude of the tension in the shaft cannot exceed T_max, so we have |T| ≤ T_max.
Substituting the expression for tension T = -mω^2r, we get |mω^2r| ≤ T_max.
Dividing both sides by mr, we get |ω^2| ≤ T_max/mr.
Taking the square root of both sides, we get |ω| ≤ √(T_max/mr).
Therefore, the maximum angular velocity at which the mass can be attached without breaking the shaft is ω_max = √(T_max/mr).
The tension in the shaft is equal in magnitude and opposite in direction to the centripetal force. Therefore, the tension in the shaft is T = -mω^2r.
In order to find the maximum angular velocity ω_max at which the mass can be attached without breaking the shaft, we need to consider the maximum tension that the shaft can withstand.
Let's assume that the maximum tension the shaft can withstand is T_max. This means that the magnitude of the tension in the shaft cannot exceed T_max, so we have |T| ≤ T_max.
Substituting the expression for tension T = -mω^2r, we get |mω^2r| ≤ T_max.
Dividing both sides by mr, we get |ω^2| ≤ T_max/mr.
Taking the square root of both sides, we get |ω| ≤ √(T_max/mr).
Therefore, the maximum angular velocity at which the mass can be attached without breaking the shaft is ω_max = √(T_max/mr).
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Consider a mass m attached at radius r from the axis of shaft which rotates with an angular velocity ω. The balancing is achieved by mounting a B kg mass at radius bfrom the axis of shaft. If the speed of shaft is doubled, then to attain perfect balance, the value of mass B should bea)halvedb)doubledc)quadrupledd)remain unchangedCorrect answer is option 'D'. Can you explain this answer?
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