Question
A 4.50-kg wheel that is 34.5 cm in diameter rotates through an angle of 13.8 rad as it slows down
uniformly from 22.0 rad/s to 13.5 rad/s. What is the magnitude of the angular acceleration of
the wheel?
A) 0.616 rad/s
2
B) 22.5 rad/s
2
C) 5.45 rad/s
2
D) 10.9 rad/s
2
E) 111 rad/s
2
uniformly from 22.0 rad/s to 13.5 rad/s. What is the magnitude of the angular acceleration of
the wheel?
A) 0.616 rad/s
2
B) 22.5 rad/s
2
C) 5.45 rad/s
2
D) 10.9 rad/s
2
E) 111 rad/s
2
Answers
ajayb
W2^2 = W1^2 + 2*alpha*theta
13.5^2 = 22.0^2 +2*alpha*13.8
alpha = (13.5^2 - 22.0^2)/2*13.8
= - 10.9 rad/s^2
So, magnitude of angular acceleration = 10.9 rad/s^2
13.5^2 = 22.0^2 +2*alpha*13.8
alpha = (13.5^2 - 22.0^2)/2*13.8
= - 10.9 rad/s^2
So, magnitude of angular acceleration = 10.9 rad/s^2