The end of a pencil of mass m and length L rests in a corner as the pencil makes an angle θ with the horizontal (x) direction (see the figure). If the pencil is released, it will rotate about point O with an acceleration α1. Suppose the length of the pencil is increased by a factor of 4 but its mass is kept the same; the pencil would then have an angular acceleration α2 when released. Find the ratio of α1/α2.
To find the ratio of α1/α2, we need to consider the rotational motion of the pencil.
Let's start by analyzing the original situation, where the pencil has mass m and length L. When released, it will rotate about point O with an acceleration α1.
We can use the concept of torque to relate the mass, length, and acceleration. Torque (τ) is given by the product of the force applied perpendicular to the rotational axis and the distance from the axis:
τ = F * r
In this case, the force causing the rotation is the weight of the pencil (mg) acting perpendicular to the axis. The distance from the axis to the center of mass of the pencil is L/2.
τ1 = (mg) * (L/2)
We also know that torque is related to angular acceleration (α) and moment of inertia (I) by the equation:
τ = I * α
The moment of inertia for a slender rod rotating about an axis through one end is given by:
I1 = (1/3) * m * L^2
Substituting these values into the torque equation:
(1/3) * m * L^2 * α1 = (mg) * (L/2)
Simplifying the equation:
α1 = (3g) / (2L)
Now, let's consider the situation where the length of the pencil is increased by a factor of 4, but the mass is kept the same.
The new length of the pencil is L' = 4L.
The moment of inertia for the new pencil can be calculated using the same formula:
I2 = (1/3) * m * (4L)^2 = (16/3) * m * L^2
The torque equation for this scenario becomes:
(16/3) * m * L^2 * α2 = (mg) * (4L/2)
Simplifying the equation:
α2 = (3g) / (8L)
Now, we can find the ratio of α1/α2:
α1/α2 = [(3g) / (2L)] / [(3g) / (8L)]
α1/α2 = (3g / 2L) * (8L / 3g)
α1/α2 = 16/6 = 8/3
Therefore, the ratio of α1/α2 is 8/3.