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A slider sliding at 10 cm/s on a link which is rotating at 60 rpm is subjected to Coriolis acceleration of magnitude
- a)40 π2 cm/s2
- b)0.4 π cm/s2
- c)40 π cm/s2
- d)4 π2 cm/s2
Correct answer is option 'C'. Can you explain this answer?
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A slider sliding at 10 cm/s on a link which is rotating at 60 rpm is s...
Coriolis acceleration
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A slider sliding at 10 cm/s on a link which is rotating at 60 rpm is s...
The Coriolis acceleration experienced by the slider can be calculated using the formula:
a_C = 2*v_L*omega_L
Where:
v_L = velocity of the slider = 10 cm/s = 0.1 m/s
omega_L = angular velocity of the link = 60 rpm = 6.28 rad/s (1 revolution = 2*pi radians)
Substituting the values, we get:
a_C = 2*0.1*6.28
a_C = 1.256 m/s^2
Therefore, the magnitude of the Coriolis acceleration experienced by the slider is 1.256 m/s^2 or approximately 1.3 m/s^2 (rounded off to one decimal place). It is not 40 as given in the options.
a_C = 2*v_L*omega_L
Where:
v_L = velocity of the slider = 10 cm/s = 0.1 m/s
omega_L = angular velocity of the link = 60 rpm = 6.28 rad/s (1 revolution = 2*pi radians)
Substituting the values, we get:
a_C = 2*0.1*6.28
a_C = 1.256 m/s^2
Therefore, the magnitude of the Coriolis acceleration experienced by the slider is 1.256 m/s^2 or approximately 1.3 m/s^2 (rounded off to one decimal place). It is not 40 as given in the options.
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A slider sliding at 10 cm/s on a link which is rotating at 60 rpm is subjected to Coriolis acceleration of magnitudea)40 π2 cm/s2b)0.4 π cm/s2c)40 π cm/s2d)4 π2cm/s2Correct answer is option 'C'. Can you explain this answer?
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