Question

A 0.250-kg wooden rod is 1.35 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 1.13 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

The moment of inertia of the rod will be:

I = 1/3 m L^2 = 1/3 x 0.25 x 1.35^2 = 0.152 kg-m^2

we know that, Torque = I alpha (alpha is angular acc)

When the rod is released the only force that acts on its gravity and through its COM, so location of com is

x = L/2

torque at this point is, F = mg x

I alpha = m g x

[1/3 m L^2 (alpha) = m g L/2 => alpha = 3 g / 2 L ]

alpha = m g x/I = 0.25 x 9.8 x (1.35/2)/0.152 = 10.889 rad/s^2

alpha = 10.889 rad/s^2

B)linear acceleration would be:

a = alpha R = 10.88 x 1.13 = 12.304 m/s^2

Hence, a = 12.304 m/s^2

C)for the given condition,

[1/3 m L^2 (alpha) = m g L/2 => alpha = 3 g / 2 L ]

X = 2/3 L

X = 2 x 1.35 / 3 = 0.900 m

Hence, X = 0.900 m