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? What is the distance rn between the point of application of n⃗ and the axis of rotation? What is the distance rw between the point of application of w⃗ and the axis?

Answer 1

Here, counterclockwise torques is positive end hence torque due to weight about the pivot point is negative as it is clockwise.  

The expression of toque, acting on pencil is given as: T =- (1/2) mgLsin(theta)

Also, the expression of torque in terms of angular acceleration is given as:

T=I a

Where I is moment of inertia of pencil and a is its angular acceleration

The moment of inertia of the pencil about one end is given as: I =(1/3)mL^2

The equation now becomes : Ia = (1/3)maL^2

Equating two expressions for torque, we get

- (1/2) mgL sin(theta) = (1/3) maL^2

a = - 3g sin(theta) /2L = - 3*9.8*sin (10)/2*0.15

So angular acceleration is a = - 17 rad/s^2

The distance rn between the point of application of n and the axis of rotation is the normal Force acting at the normal Force only, therefore the distance between them rn is zero.

The gravity acts at exactly half of the length of the pencil. Hence the distance between the weight and point of the application is

rw = Lcos(theta)/2= 0.15*cos(10)/2= 0.07386m =0.074 m

Therefore, rn= 0 and rw = 0.074 m