A 250 gg , 25-cm-diameter plastic disk is spun on an axle through its center by an electric motor. What torque must the motor supply to take the disk from 0 to 1800 rpm in 4.3 s ?

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A 250 gg , 25-cm-diameter plastic disk is spun on an axle through its center by an electric motor.  What torque must the motor supply to take the disk from 0 to 1800 rpm in 4.3 s ?

raefx88y · Accepted Answer

Mass of disk, m=250g⇒0,25kg  Diameter of disk, α=25cm⇒0,25m  Initial angular velocity, ωi=0rads  Final angular velocity, ωf=1800RPM  =2π×180060rads  188,5rads  Time, t=4,3s  Angular acceleration , α=ωf−ωit  ∴α=188,5−04,3⇒43,84rads2  Moment of inertia od disk about its centre : I=12mr2  Where r=α2  ∴I=12×0,25×(0,252)2  I=1,95×10−3kg−m2  Using relation: torque, τ=Iα  ∴τ=1,95×10−3×43,84  τ=8,56×10−2N−m

Mollie Nash · Accepted Answer

The point where the individual fields cancel cannot be in the region between the shells since the shells have opposite-signed charges. It cannot be inside the radius R of one of the shells since there is only one field contribution there (which would not be canceled by another field contribution and thus would not lead to zero net fields). We note shell 2 has a greater magnitude of charge (|σ2|A2) than shell 1, which implies the point is not to the right of shell 2 (any such point would always be closer to the larger charge and thus no possibility for cancellation of equal-magnitude fields could occur). Consequently, the point should be in the region to the left of shell 1 (at a distance r&amp;gt;R1 from its center); this is where the condition   E1=E2⇒|q1|4πϵ0r2=|q1|4πϵ0(r+L)2  σ1A14πϵ0r2=|σ2|A24πϵ0(r+L)2  Using the fact that the area of a sphere is A=4πR2, this condition simplifies to :   r=L(R2R1)σ2σ1−1=3.3cm  We note that this value satisfies the requirement r&amp;gt;R1. The answer, then, is that the net field vanishes at x=−r=−3.3cm.

alenahelenash · Accepted Answer

Thanks for the answers, I got an excellent rating.

A 250 gg , 25-cm-diameter plastic disk is spun on

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