Question

When a carpenter shuts off his circular saw, the 10.0-inch-diameter blade slows from 4987 rpm to zero in 2.25 s .

What is the distance traveled by a point on the rim of the blade during the deceleration?

What is the magnitude of the net displacement of a point on the rim of the blade during the deceleration?

Answer 1

Ans:-

Given data:- d= 10inch, w= 4987rev/min, t = 2.25s

w= 4987rev/min*2π rads / rev* 1min/60s = 521.97rad/s

α = ∆w/∆t = - 521.97rads/3.00s = -145rad/s^2

by using kinematics equation for rotational motion

θ = w*t + ½*αt^2

= 521.97*2.25 - ½*145rad/s^2*(2.25)^2

Θ=1174.43 – 367.03

Θ = 807.4rads

S= θ*r = 807.4rads*5in*0.0254m/in = 102.54m

Now find net displacement

Θ = 807.4

θ/2π = 128.57revolutions exactly

so the net is ½ a revolution and therefore the net displacement in one diameter

d= 10in *0.0254m/in = 0.254m = 25.4cm