Q1:A circular saw blade of diameter 0.250 m starts from rest. In a time interval

Question

Q1:A circular saw blade of diameter 0.250 m starts from rest. In a time interval of 5.80 s it accelerates with constant angular acceleration to an angular velocity of 134 rad/s .

a.Find the angular acceleration.

b.Find the angle through which the blade has turned.

Q2:A wheel is turning about an axis through its center with constant angular acceleration. Starting from rest, at t=0, the wheel turns through 7.50 revolutions in t= 12.5 s . At 12.5 s the kinetic energy of the wheel is 39.0 J .

a.For an axis through its center, what is the moment of inertia of the wheel?

Explanation / Answer

Q1: t = 5.8 sec
o = 0
f = 134 rev/sec
a = ?
Convert the linear velocity equation into a circular velocity equation.
Vf = Vo + at (linear)
f = o + at (circular)
134 rev/sec = a (5.8 sec)
(a) a = 23.10 rev/sec²
= ?
Convert another linear velocity equation into a circular velocity equation.
Vf² = Vo² + 2aD
f² = o² + 2a
(134 rev/sec)² = 0 + 2 (23.10 rev/sec²)
17956 rev²/sec² = 46.20 rev/sec² ()
= 388.66 rev
= 388.66 rev (360º/1 rev) = 139917.6º
(b) = 139917.6º is the distance the blade has turned

Q2:

2 . 7.50 = 0.5 . a . 12.5^2

a = 0.60 rad/s/s

w^2 = 2 .a . d = 2 . 0.60 . 2 . 7.5 = 56.52

(a) I = 2 . KE / w^2 = 2*39.0 / 56.52 = 1.38 kgm^2

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