Question

A 0.180-kg wooden rod is 1.60 m long and pivots at one end. It is held horizontally and then released.

Part A : What is the angular acceleration of the rod after it is released? Express your answer to three significant figures and include appropriate units.

Part B: What is the linear acceleration of a spot on the rod that is 0.560 m from the axis of rotation? Express your answer to three significant figures and include appropriate units.

Part C: 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

Part A.

Angular acceleration of the rod will be given by:

/I

= torque applied to the rod = F*R = m*g*(L/2)

I = moment of inertia of rod about it's one end = m*L^2/3

So,

/I = m*g*(L/2)/(m*L^2/3)

= 3*g/(2*L)

L = length of rod = 1.60 m

So,

= 3*9.81/(2*1.60) = 9.20 rad/sec^2

Part B.

Linear acceleration is given by:

a = *R

R = radius of spot's motion about the axis of rotation = 0.560 m

a = 9.20*0.560

a = 5.15 m/sec^2

Part C.

die will just begin to to separate from the rod when it's linear acceleration will be equal to gravitational acceleration.

a = g

*R1 = g

R1 = g/

R1 = 9.81/9.20 = 1.066 m

R1 = location along the rod where die just begins to separate = 1.07 m

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