Question

A 0.240-kg wooden rod is 1.25 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.

What is the linear acceleration of a spot on the rod that is 0.388 m from the axis of rotation?

Express your answer to three significant figures and include appropriate units.

At what location along the rod should a die be placed so that the die just begins to separate from the rod as it falls?

Express your answer to three significant figures and include appropriate units.

Answer 1

Solution:

Mass of the wooden rod = 0.240 kg = M

Length = L = 1.25 m

Weight of the rod acts through its center , vertically downwards and so the Torque due to this gravity force is

= r /2 * mg sin 90 = (1.25/2)(0.240 * 9.8) = 1.47 N m

Moment of Inertia of the rod pivoted at one end = I = 1/3 ML^2= 1/3 * 0.240 * (1.25)^2

= 0.125

torque = = I

=> = / I = (1.47) / (0.125) = 11.76 rad/s^2 =11.8 rad/s/s (3 significant figures)

b) At a distance of 0.388 m , the acceleration a = 0.388 * 11.76 = 4.56 m/s^2

c) For the die to just fall, let location = R

g = R

=> R = g / = 9.8 / 11.76 = 0.833 m =8.33 x 10^-1 m