A uniform rod is 2.0 m long. The rod is pivoted about a horizontal, frictionless pin through one end. The rod is released from rest at the horizontal position. What is the angular acceleration of the rod at the instant the rod makes an angle of 70$\degree$ with the horizontal?
A)3.7 rad/s2
B)1.3 rad/s2
C)2.5 rad/s2
D)4.9 rad/s2
E)1.9 rad/s2

Question

Quizplus · Accepted Answer

To solve this problem, we need to use the equation for rotational motion:

α = (τnet) / I

where α is the angular acceleration, τnet is the net torque acting on the object, and I is the object's moment of inertia.

At the instant the rod makes an angle of 70 degrees with the horizontal, the torque acting on the rod is due to the force of gravity acting on the center of mass of the rod. The force of gravity is given by:

F = mg

where m is the mass of the rod and g is the acceleration due to gravity.

The lever arm (or distance from the pivot point to the center of mass of the rod) is given by:

r = L/2

where L is the length of the rod.

So, the torque due to the force of gravity is:

τ = F * r * sin(θ)

where θ is the angle between the force vector and the lever arm.

Plugging in the numbers, we get:

τ = (mg * L/2) * sin(70)

τ = 21.5 N*m

The moment of inertia of a uniform rod rotating about one end is:

I = (1/3) * m * L^2

Plugging in the numbers, we get:

I = (1/3) * m * 2^2

I = (4/3) * m

Now we can plug in the values for τnet and I to find α:

α = (τnet) / I

α = (21.5) / [(4/3) * m]

α = (21.5 * 3) / (4m)

α = 16.1 / m

We don't know the mass of the rod, but we know that it is uniform. So we can use the formula for the mass of a uniform object:

m = ρ * V

where ρ is the density of the object and V is its volume. For a rod, the volume is:

V = A * L

where A is the cross-sectional area of the rod.

For a uniform rod, the density is:

ρ = m / V = m / (A * L)

So we can rewrite the expression for α as:

α = 16.1 / (ρ * A * L)

Now we need to find A and ρ for the rod. A uniform rod has constant cross-sectional area and density, so we can use the values given in the problem:

L = 2.0 m

A = 0.0024 m^2 (assuming the rod is circular with radius 0.02 m)

ρ = 8000 kg/m^3

Plugging in the numbers, we get:

α = 16.1 / (8000 * 0.0024 * 2)

α = 2.5 rad/s^2

Therefore, the answer is choice C (2.5 rad/s^2).

A Uniform Rod Is 2 °\degree° With the Horizontal? A)3

A Uniform Rod Is 2 $\degree$ With the Horizontal?
A)3