Question

What is the angular acceleration of the pencil when it makes an angle of 10.0 degrees with the vertical?

Answer 1

r:  If they give you the net torque and the moment of inertia already, it's pretty easy. 

torque = I * alpha, where I is moment of inertia, alpha is angular acceleration. 

Thus, 0.098 / 0.000075 = 1306.666... rad / s^2 

The angular acceleration remains constant, but the angular velocity at that point (if you want it) is 21.36 rad / s

Answer 2

The angular acceleration of the pencil when it makes an angle of 10.0 degrees with the vertical is 1306.6 rad / s^2

Explanation:

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?  Express your answer in radians per second squared.

When the pencil is balanced, the net torque acting on the pencil is zero. Also when we released the pencil and begin to fall, net torque acts on the pencil because of the pencil weight.

We know about the relation between torque and angular acceleration. Therefore the angular acceleration of the pencil is

The pencil length L = 15 cm and mass m have an angle (θ) 10­­0 with the verticle.  Then the torque(τ) acting on the pencil due to its weight (W) is

$\tau = r xF\\\tau =\frac{L}{2} W sin \theta\\\tau =\frac{L}{2} mg sin 10^o$

$\tau =\frac{0.15}{2} m*10* sin 10^o\\\tau = 0.075  m*10* sin 10^o\\\tau = 0.130m$

Then the angular acceleration is

$\tau = I \alpha = F r\\\tau = 0.130m = F r\\\alpha = \frac{0.098}{0.000075}  = 1306.6 rad / s^2$

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