Question

What is the angular acceleration of the pencil when it makes an angle of 10.0 degrees with the vertical? express your answer numerically in radians per second squared to three significant figures?

Answer 1

`\alpha = (3gsin10)/(2L)`

Step-by-step explanation:

As we know by Newton's law

`\tau = I\alpha`

here torque is due to the weight of the pencil which will act on the center of the pencil

so here we will have

`\tau = mg(L)/(2)sin\theta`

now we know that the moment of inertia of the pencil is given as

`I = (mL^2)/(3)`

now we will have

`mg(L)/(2)sin\theta = (mL^2)/(3)\alpha`

`\alpha = (3g)/(2L)sin\theta`

for 10 degree angle we have

`\alpha = (3gsin10)/(2L)`