How Does Torque Apply to a Rotating Rod on a Frictionless Surface?

[Ethidium]

Homework Statement

A uniform rod is released on a friction-less horizontal surface from rest. Initially, the rod makes an angle θ = 60° with the horizontal. What is the acceleration of the center of mass of the rod, just after release?

Homework Equations

Torque = Moment of Inertia × Angular acceleration
Angular acceleration α = dω/dt = d²θ/dt²
Mg - N = Ma, where M is the mass of the rod, a is the acceleration about the rod's center of mass and N is the normal reaction.

The Attempt at a Solution

The only two forces acting on the rod: the normal reaction by the surface and the weight of the rod.
Finding torque about the point of contact with the surface, we get: (M is mass of the rod, L is it's length)
Mg(L/2)cosθ = ⅓(ML²)α ..... (1)
Also if we balance forces, we get: (a is the acceleration of centre of mass of the rod, N is the normal reaction)
Mg - N = Ma ...(2)

Now I realize that the axis of rotation is not fixed and thus wonder how to go about using equation (1). I also cannot use a condition of pure rotation such as a = rα. I'm guessing I will get a second order differential equation by virtue of the fact that α = d²(θ)/dt².
So, Mg(L/2)cosθ = ⅓ML²(d²θ/dt²)
⇒ secθ d²θ/dt² = (3g)/(2L)
Am I required to solve this differential equation?

 

[TSny]

Welcome to PF!

Finding torque about the point of contact with the surface, we get: (M is mass of the rod, L is it's length)
Mg(L/2)cosθ = ⅓(ML²)α ..... (1)
Also if we balance forces, we get: (a is the acceleration of centre of mass of the rod, N is the normal reaction)
Mg - N = Ma ...(2)

Now I realize that the axis of rotation is not fixed and thus wonder how to go about using equation (1).

Equation (1) is not valid, as you suspect. The point of contact is itself accelerating. However, it can be shown that ##\tau = I \alpha## is valid about the center of mass even though the center of mass is accelerating! Love that center of mass.

You will also need to find a kinematic relationship between ##\alpha## and the acceleration of the center of mass.

You will not need to solve a differential equation. You just need to find the acceleration of the center of mass immediately after release.

 

[Ethidium]

Welcome to PF!

Equation (1) is not valid, as you suspect. The point of contact is itself accelerating. However, it can be shown that ##\tau = I \alpha## is valid about the center of mass even though the center of mass is accelerating! Love that center of mass.

You will also need to find a kinematic relationship between ##\alpha## and the acceleration of the center of mass.

You will not need to solve a differential equation. You just need to find the acceleration of the center of mass immediately after release.

Thanks a lo

Welcome to PF!

Equation (1) is not valid, as you suspect. The point of contact is itself accelerating. However, it can be shown that ##\tau = I \alpha## is valid about the center of mass even though the center of mass is accelerating! Love that center of mass.

You will also need to find a kinematic relationship between ##\alpha## and the acceleration of the center of mass.

You will not need to solve a differential equation. You just need to find the acceleration of the center of mass immediately after release.

Welcome to PF!

Equation (1) is not valid, as you suspect. The point of contact is itself accelerating. However, it can be shown that ##\tau = I \alpha## is valid about the center of mass even though the center of mass is accelerating! Love that center of mass.

You will also need to find a kinematic relationship between ##\alpha## and the acceleration of the center of mass.

You will not need to solve a differential equation. You just need to find the acceleration of the center of mass immediately after release.

Thanks a lot. I will try finding the kinematic relationship.
I was also confused about another thing (it is a bit silly). The point of contact is accelerating horizontally right? But how exactly is it accelerating if there is no net horizontal force on it? The only forces acting on it are vertical.

 

[PeroK]

Thanks a loThanks a lot. I will try finding the kinematic relationship.
I was also confused about another thing (it is a bit silly). The point of contact is accelerating horizontally right? But how exactly is it accelerating if there is no net horizontal force on it? The only forces acting on it are vertical.

The bar has internal forces between its molecules. There can produce motion in a direction for which there is no external force.

For example, a rotating bar may have no external forces, but each point in the bar is executing circular motion.

 

[Ethidium]

The bar has internal forces between its molecules. There can produce motion in a direction for which there is no external force.

For example, a rotating bar may have no external forces, but each point in the bar is executing circular motion.

Ahhh right. Thanks