Answered

An electrical motor spins at a constant 2695.0 rpm. If the armature radius is 7.165 cm, what is the acceleration of the edge of the rotor?
A) 28.20 m/s^2

B) 572,400 m/s^2

C) 281.6 m/s^2

D) 5707 m/s^2

Answer :

Answer:

Acceleration will be [tex]5706.77rad/sec^2[/tex]

So option (D) will be correct answer

Explanation:

We have given angular speed of the electrical motor [tex]\omega =2695rpm[/tex]

We have to change this angular speed in rad/sec for further calculation

So [tex]\omega =2695rpm=2695\times \frac{2\pi }{60}=282.2197rad/sec[/tex]

Armature radius is given r = 7.165 cm = 0.07165 m

We have to find the acceleration of edge of motor

Acceleration is given by [tex]a=\omega ^2r=282.2197^2\times 0.07165=5706.77rad/sec^2[/tex]

So acceleration will be [tex]5706.77rad/sec^2[/tex]

So option (D) will be correct answer

Cricetus

The acceleration of the edge of rotor will be "5707 rad/s²".

Given:

Radius,

  • r = 7.165 cm

or,

          = 0.07165 m

Angular speed,

  • [tex]\omega[/tex] = [tex]2695.0 \ rpm[/tex]

or,

          = [tex]2695\times \frac{2 \pi}{60}[/tex]

          = [tex]282.2197 \ rad/sec[/tex]

Now,

The acceleration will be:

→ [tex]a = \omega^2 r[/tex]

     [tex]= 282.2197^2\times 0.07165[/tex]

     [tex]= 5706.77 \ rad/sec^2[/tex]

Thus the above answer is right.

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