Question

A circular saw blade accelerates from rest to an angular speed of 3620 rpm in 6.30 revolutions. A) Find the torque exerted on the saw blade, assuming it is a disk of radius 17.2 cm and mass 0.735 kg . B) Explain your response. C) Find the angular speed of the blade after 3.15 revolutions.

Answer 1

given
m = 0.735 kg
r = 17.2 cm = 0.172 m
wi = 0
wf = 6260 rpm = 6260*2*pi/60 = 656 rad/s
theta = 6.30 revolutions = 6.30*2*pi = 39.6 rad

a) angular acceleration of the disk, alfa = (wf^2 - 2i^2)/(2*theta)

= (656^2 - 0^2)/(2*39.6)

= 5434 rad/s^2

moment of inertia of the disk, I = 0.5*m*r^2

= 0.5*0.735*0.172^2

= 0.0109 kg.m^2

Torque = I*alfa

= 0.0109*5434

= 59.2 N.m <<<<<<<-----Answer

B) I am assuming the disk rotates with constant angular acceleration.

ans I ams assuming the torque is also constant.

C) theta = 3.15 rev = 3.15*2*pi = 19.8 rad

wf^2 - wi^2 = 2*alfa*theta

wf = sqrt(2*alfa*theta) (since wi = 0 )

= sqrt(2*5434*19.8)

= 464 rad/s

= 464*60/(2*pi)

= 4430 rpm <<<<<<<<<<<-------------------Answer