A combination lock has a 1.1-cm-diameter knob that is part of the dial you turn

Question

A combination lock has a 1.1-cm-diameter knob that is part of the dial you turn to unlock the lock. To turn that knob, you grip it between your thumb and forefinger with a force of 4.0 N as you twist your wrist. Suppose the coefficient of static friction between the knob and your fingers is  0.80.

Part A answer is 3.520 x10-2 Nm (What is the most torque you can exert on the knob without having it slip between your fingers?)

Part B If the maximum torque causes the knob to rotate at 15RPM after 30s of being constantly applied, what is the knob's moment of inertia? Assume the knob starts at rest.

**Answer is not 0.92 x 10^-3 kgxm^2

Explanation / Answer

Torque = Force× radius

Force =cofficient of friction ×normal reaction

Force=4×0.8=3.2

Radius =1.1×10^-2/2

Torque =3.2×1.1×10^-2/2=1.76×10^-2Nm

Torque by thumb =1.76×10^-2Nm

Torque by forefinger =1.76×10^-2Nm

Max .torque =(1.76+1.76)×10^-2=3.52×10^-2Nm

Part B ) Torque = I× angular acceleration (equation 1)

W=wo+angular acceleration ×t

W=15Rpm=15×2×pi/60=pi/2=6.28rad/sec

6.28=0+angular acceleration ×30

Angular acceleration =6.28/30=0.20933rad/sec^2

Using equation 1 we get 3.52×10^-2=0.20933I

I=16.815×10-2Kg

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