Question

A grinding wheel 0.37 m in diameter rotates at 2400 rpm.(A) Calculate its angular velocity in rad/s.(B) What is the linear speed of a point on the edge of the grinding wheel? (C) What is the acceleration of a point on the edge of the grinding wheel?

Answer 1

Answer:

A) Angular velocity = 251.36 rad/s

B) Linear speed = 46.50 m/s

C) Acceleration = 11687.8 m/s^2

Explanations:

A) Angular velocity, w = 2400 rpm

Note that:

1 rpm = (2π)/60

Therefore:

2400 rpm = (2π/60) x 2400 rad/s

2400 rpm = 80π rad/s

2400 rpm = 80 x 3.142

2400 rpm = 251.36 rad/s

The angular velocity, w = 251.36 rad/s

B) The relationship between linear speed(v) and angular velocity(w) is:

v = wr

The diameter, d = 0.37m

Radius, r = d/2

r = 0.37/2

r = 0.185m

Substituting r = 0.185m and w = 251.36 rad/s into v = wr:

v = 251.36 x 0.185

v = 46.50 m/s

C) The acceleration a point on the edge of the grinding wheel is the centripetal acceleration, and is given by the formula:

$\begin{gathered} a\text{ = }\frac{v^2}{r} \\ a\text{ = }\frac{46.50^2}{0.185} \\ a\text{ = }\frac{2162.25}{0.185} \\ a\text{ = }11687.8m/s^2 \end{gathered}$