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2 years ago

13

An electrical motor spins at a constant 2662.0 rpm. If the armature radius is 6.725 cm, what is the acceleration of the edge of

the rotor? O 524,200 m/s O 29.30 m/s O292.7 m/s O 5226 m/s2

1 answer:

Gemiola [76]2 years ago

3 0

Answer:

18.73 m/s^2

Explanation:

f = 2662 rpm = 2662 / 60 rps

r = 6.725 cm = 0.06725 m

Acceleration, a = r w

a = r x 2 x pi x F

a = 0.06725 × 2 × 3.14 × 2662 / 60

a = 18.73 m/s^2

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If the disk had twice the radius and twice the mass

The new moment of inertia

$I'=\dfrac{1}{2}\times2M\times(2R)^2$

$I'=8I$

We know,

The torque is

$\tau=F\times R$

We need to calculate the initial rotation acceleration

Using formula of acceleration

$\alpha=\dfrac{\tau}{I}$

Put the value in to the formula

$\alpha=\dfrac{F\times R}{\dfrac{1}{2}MR^2}$

$\alpha=\dfrac{2F}{MR}$

We need to calculate the new rotation acceleration

Using formula of acceleration

$\alpha'=\dfrac{\tau}{I'}$

Put the value in to the formula

$\alpha=\dfrac{F\times R}{8\times\dfrac{1}{2}MR^2}$

$\alpha=\dfrac{2F}{8MR}$

$\alpha=\dfrac{\alpha}{8}$

Rotation speed is same.

We need to calculate the time

Using formula angular velocity

$\Omega=\omega'$

$\alpha\time t=\alpha'\times t'$

Put the value into the formula

$\alpha\times120=\dfrac{\alpha}{8}\times t'$

$t'=960\ sec$

$t'=16\ min$

Hence, The time is 16 min.

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