Question

The blade of a table saw has a diameter of 25.4 cm and rotates with a frequency of 750. What is the magnitude of the centripetal acceleration at the edge of the blade?

Answer 1

Дано:

D = 25,4 см = 0,254 м

R = D/2 = 0,127 м

ω = 750 рад/с

___________

aₙ - ?

Центростремительное ускорение

aₙ = ω²·R

aₙ = 750² · 0,127 ≈ 71 400 м/с²

Given:

D = 25.4 cm = 0.254 m

R = D / 2 = 0.127 m

ω = 750 rad / s

___________

aₙ -?

Centripetal acceleration

aₙ = ω² · R

aₙ = 750² · 0.127 ≈ 71,400 m / s²

Answer 2

To find the magnitude of the centripetal acceleration at the edge of the blade, we can use the formula for centripetal acceleration:

a = (v^2) / r

Where: a is the centripetal acceleration, v is the linear velocity, and r is the radius.

First, let's find the linear velocity of a point on the edge of the blade. The circumference of a circle is given by:

C = 2πr

Given that the diameter of the blade is 25.4 cm, the radius (r) is half of that:

r = 25.4 cm / 2 = 12.7 cm = 0.127 m

The linear velocity (v) can be calculated by multiplying the circumference (C) with the frequency (f):

v = C * f

Substituting the values:

v = (2πr) * f = (2π * 0.127 m) * 750

Now we can calculate the centripetal acceleration (a) using the formula:

a = (v^2) / r

a = ((2π * 0.127 m) * 750)^2 / 0.127 m

Evaluating this expression will give us the magnitude of the centripetal acceleration at the edge of the blade.