Project ForgeA mobile phone has a small motor with an eccentric mass used for vibrator mode, resting flat on a horizontal surface between its two ends P and Q. Its center of gravity (CG) is 6 cm from end P; the motor with its eccentric mass sits a further 3 cm along (i.e. 9 cm from P), close to end Q. Given the eccentric mass=2 grams, eccentricity=2.19 mm, mass of the mobile=90 grams, g=9.81 m/s². The uniform speed of the motor in RPM for which the mobile will get just lifted off the ground at end Q is approximately
- A3000
- 3500
- C4000
- D4500
Explanation
End Q lifts off (its ground reaction → 0) when, taking moments about the far end P, the peak upward unbalance force at the motor (9 cm from P) just balances the weight's moment (weight acts at the CG, 6 cm from P): W×6 = Fmax×9. W=mg=0.090×9.81≈0.8829 N, so Fmax=0.8829×6/9≈0.5886 N. The unbalance force amplitude is Fmax=me·e·ω²: ω²=0.5886/(0.002×0.00219)≈134,384, ω≈366.6 rad/s. Converting: N=ω×60/(2π)≈3501 rpm, matching option (B), 3500.
Source: GATE ME 2015 (Session 1) (2015)
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