Question

A grinding wheel 0.35 m in diameter rotates at 2600 rpm .

Answer 1

It is calculated that a)The angular velocity of the wheel is 272.13 rad/s,

b)On the edge of the grinding wheel, the linear speed is 47.62 m/s,

and c) On the edge of the grinding wheel, the acceleration is 12958.08 m/s².

Calculation of angular velocity, linear speed & acceleration:

Provided that,

the diameter of the wheel = 0.35 m

So, the radius, r = 0.35/2 = 0.175 m

As 1 revolution = 2π rad

(a)the angular velocity, ω = 2600 rpm = $\frac{2600 * 2\pi }{60}$ rad/s

⇒ω = 272.13 rad/s

So, the angular velocity is 272.13 rad/s.

(b)The linear speed, v = r * ω

⇒v = 0.175 * 272.13

⇒v= 47.62 m/s

(c)The angular acceleration, $a=\frac{v^{2} }{r}$

$a = \frac{(47.62)^{2} }{0.175}$

⇒$a$ = 12958.08 m/s²

Learn more about angular velocity here:

brainly.com/question/13649539

#SPJ1