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A mobile phone has a small motor with an eccentric mass used for vibrator mode. The location of the eccentric mass on motor with respect to center of gravity (CG) of the mobile and the rest of the dimension of the mobile phone are shown. The mobile is kept on a flat horizontal surface.
Given in addition that the eccentric mass = 2 grams, eccentricity = 2.19 mm, mass of the mobile = 90 grams, g= 9.81 m/s2. Uniform speed of the motor in RPM for which the mobile will get just lifted off the ground at the end Q is approximately
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- 3000
- 3500
- 4000
- 4500
Correct Option: B
When lifted from ground at Q
Reaction = 0
∴ taking moments about ‘p’ and equating to 0
.09 × .06 = mr ω2 × .09
9.01 × .09 × .06 = .002 × 2.19 × 10-3 × ω2 × .09
⇒ ω = 366.50 rad/s
= 3500 rpm
