Mary Buchanan

2022-01-14

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 ?

What torque must the motor supply to take the disk from 0 to 1800 rpm in 4.3 s ?

raefx88y

Beginner2022-01-15Added 26 answers

Mass of disk, $m=250g\Rightarrow 0,25kg$

Diameter of disk,$\alpha =25cm\Rightarrow 0,25m$

Initial angular velocity,$\omega}_{i}=0ra\frac{d}{s$

Final angular velocity,${\omega}_{f}=1800RPM$

$=\frac{2\pi \times 1800}{60}ra\frac{d}{s}$

$188,5ra\frac{d}{s}$

Time,$t=4,3s$

Angular acceleration ,$\alpha =\frac{{\omega}_{f}-{\omega}_{i}}{t}$

$\therefore \alpha =\frac{188,5-0}{4,3}\Rightarrow 43,84ra\frac{d}{{s}^{2}}$

Moment of inertia od disk about its centre :$I=\frac{1}{2}m{r}^{2}$

Where$r=\frac{\alpha}{2}$

$\therefore I=\frac{1}{2}\times 0,25\times {\left(\frac{0,25}{2}\right)}^{2}$

$I=1,95\times {10}^{-3}kg-{m}^{2}$

Using relation: torque,$\tau =I\alpha$

$\therefore \tau =1,95\times {10}^{-3}\times 43,84$

$\tau =8,56\times {10}^{-2}N-m$

Diameter of disk,

Initial angular velocity,

Final angular velocity,

Time,

Angular acceleration ,

Moment of inertia od disk about its centre :

Where

Using relation: torque,

Mollie Nash

Beginner2022-01-16Added 33 answers

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 $\left(\left|{\sigma}_{2}\right|{A}_{2}\right)$ 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 $r>{R}_{1}$ from its center); this is where the condition

$E}_{1}={E}_{2}\Rightarrow \frac{\left|{q}_{1}\right|}{4\pi {\u03f5}_{0}{r}^{2}}=\frac{\left|{q}_{1}\right|}{4\pi {\u03f5}_{0}{(r+L)}^{2}$

$\frac{{\sigma}_{1}{A}_{1}}{4\pi {\u03f5}_{0}{r}^{2}}=\frac{\left|{\sigma}_{2}\right|{A}_{2}}{4\pi {\u03f5}_{0}{(r+L)}^{2}}$

Using the fact that the area of a sphere is$A=4\pi {R}^{2}$ , this condition simplifies to :

$r=\frac{L}{\left(\frac{{R}_{2}}{{R}_{1}}\right)\sqrt{\frac{{\sigma}_{2}}{{\sigma}_{1}}}-1}=3.3cm$

We note that this value satisfies the requirement$r>{R}_{1}$ . The answer, then, is that the net field vanishes at $x=-r=-3.3cm$ .

Using the fact that the area of a sphere is

We note that this value satisfies the requirement

alenahelenash

Skilled2022-01-23Added 488 answers

Thanks for the answers, I got an excellent rating.

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