Suppose that you are holding a pencil balanced on its point. If you release the

Question

Suppose that you are holding a pencil balanced on its point. If you release the pencil and it begins to fall, what will be the angular acceleration when it has an angle of 10.0 degrees from the vertical? There are two reasonable approximations to consider for the pencil in this problem: a cylinder and a thin rod. However, in this problem we will treat the pencil as a uniform thin rod of length 15.0 cm and mass 10.0g .What is the angular acceleration of the pencil when it makes an angle of 10.0 degrees with the vertical?

Known:
m=10g
rn-=0cm
rw=7.5cm
=10 degrees

g=9.8 m/s2

Explanation / Answer

length of the pencil L = 0.15 m

= 10.0o

net torque = mg(L/2)sin = I, where I = mL2/3

= 3gsin/(2L) = 17.0 rad/s2

Torque and angular acceleration are both vectors, their sign mean directions. In this problem if we define the direction of increasing angle as positive, then torque and angular acceleration are both negative.

So it could also be - 17.0 rad/s^2, depending on how you orient your axis.

Hope that helps, good luck!

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