Complete Question.
A. 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?
B. What is the distance rn between the point of application of n⃗ and the axis of rotation?
C. What is the distance rw between the point of application of w⃗ and the axis? enter your answers in meters separated by a comma?
Answer:
A. â = - 17rad/s²
B. rn = 0, meters
C. rw = 0.074, meters
Explanation:
Where;
l = length of the pencil
θ = angle between the vertical line and the pencil
â = angular acceleration.
The torque due to weight about the pivotal point is negative since it's clockwise;
t = - mg(l/2)sinØ ..... equation 1
torque with respect to angular acceleration;
t = Iâ ..... equation 2
The moment of Inertia of the pencil about one end;
I = (ml²)/3
Substituting I into equation 2;
t = [(ml²)/3]×â
Equating the equation, we have;
[(ml²)/3]×â = - mg(l/2)sinØ
â = (-3gsinØ)/2l
â = (-3*9.8*sin10°)/(2*0.15)
Angular acceleration, â = -17rad/s²
B. The normal force is acting at the normal force only, so the distance r n between the point of application of n ⃗ and the axis of rotation is zero because the axis of pencil is passing the same point of application w.
C. rn = (lcosØ)/2
rn = (0.15 * cos10°)/2
rn = (0.15 * 0.9848)/2
rn = (0.1477)/2
rn = 0.074m.
Since the gravity acts at exactly half of the length of pencil, distance r w between the point of application of w⃗ and the axis equals 0.074m.
