Question

oblem 11.14 6 of 2 A 0.190-kg wooden rod is 1.10 m long and pivots at one end It is held horizontally and then released What is the angular acceleration of the rod after it is released? Express your answer to three significant figures and include appropriate units. Value Units Part B What is the linear acceleration of a spot on the rod that is 0.836 Im from the axis of rotation? Express your answer to three significant figures and include appropriate units. Request Answer 854 PM 4/22/2018 Type here to search

Answer 1

Part A)

We know that the relation between Torque and angular accelaration is

T = I*alpha

(r*F*sin(90)) = (1/3)*M*R^2*alpha

(l/2)*m*g = (1/3)*M*l^2*alpha

(1.1/2)*0.19*9.8 = (1/3)*0.19*1.1^2*alpha

alpha = 13.36 rad/s^2

Part B)

linear accelaration is a = r*alpha = 0.836*13.36 = 11.17 m/s^2