Question

A mobile phone has a small motor with an eccentric mass used for vibrator mode. The location of the eccentric mass on the motor with respect to the center of gravity (CG) of the mobile and the rest of the dimensions 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 ${\frac{m}{s}}^2.$ Uniform speed of the motor in RPM for which the mobile will get just lifted off the ground at the end Q is approximately.

• A. 3000
• B. 3500
• C. 4000
• D. 4500

Answer 1

The correct option is B 3500


$\Sigma M_P=0$[moment about P= 0]

$mg\times 6=m_e(2.19\times {10}^{-3}({\omega }^2)\times [9]$

$90\times 9.81\times 6=2(2.19\times {10}^{-3}){\omega }^2\times 9$

$\omega =366.58\frac{rad}{s}$

also, $\omega =\frac{2\pi N}{60}$

$\therefore 366.58=\frac{2\pi \times N}{60}$
or N= 3500.61 rpm = 3500 rpm