Question

Q3 A rotor 25mm in diammeter is spinning out 200rps.Find normal component of acceleration of a point on rim . (A) 20000 m/s^2 (B) 19800m/s^2 (C) 19739 m/s^2
​

Answer 1

Answer:

c

Explanation:

c is the correct answer

Answer 2

Answer:

The normal component of acceleration of a point on the rim is 19739ms⁻².i.e.option(C).

Explanation:

The normal component of acceleration of a point on the rim is centripetal acceleration. Which is given as,

$a_c=\omega^2r$            (1)

ac=centripetal acceleration

ω=angular velocity

r=radius of the rim

From the question we have,

Frequency(f)=200 rps

Diameter=25mm=25×10⁻³m

So, the radius will be=25×10⁻³m/2

Now first we need to find angular velocity which is given as,

$\omega=2\pi f$           (2)

By putting the value of frequency in the above equation we get;

$\omega=2\pi \times 200=400\pi$          (3)

By putting the required values in equation (1) we get;

$a_c=(400\pi)^2\times \frac{25\times 10^-^3}{2}\approx19739ms^-^2$

Hence, the normal component of acceleration of a point on the rim is 19739ms⁻².i.e.option(C).