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

• Ethidium
In summary: 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 force acting on it, but may be moving in a direction for which there is no external force because of the internal forces between the molecules.In summary, a rotating bar can move in a direction for which there is no external force because of the internal forces between the molecules.
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?

Last edited:
Welcome to PF!
Ethidium said:
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 and PeroK
TSny said:
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
TSny said:
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.
TSny said:
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.

Ethidium said:
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
PeroK said:
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

## What is τ = Iα used for?

Torque, represented by τ, is a measure of the rotational force applied to an object. In this equation, I represents the moment of inertia and α represents the angular acceleration. It is most commonly used in physics and engineering to calculate the rotational motion of objects.

## How do I calculate τ = Iα?

To calculate torque using this equation, you will need to know the moment of inertia (I) and the angular acceleration (α) of the object. Then, simply multiply I by α to get the value of torque. Keep in mind that the unit for torque is Newton-meters (Nm).

## What is the difference between τ = Iα and F = ma?

While both equations involve force and acceleration, they are used for different types of motion. F = ma is used for linear motion, while τ = Iα is used for rotational motion. Additionally, F represents the net force applied to an object, while τ represents the rotational force applied to an object.

## Can τ = Iα be used for all types of rotational motion?

Yes, τ = Iα can be used for any type of rotational motion, as long as the moment of inertia and angular acceleration are known. It is a fundamental equation in rotational dynamics and is applicable in a wide range of scenarios, from simple systems like a spinning top to more complex systems like a rotating planet.

## What are some real-life applications of τ = Iα?

Torque is used in various real-life applications, such as designing car engines, calculating the forces on a rotating wind turbine, and analyzing the motion of a spinning top or gyroscope. It is also important in sports, as it helps explain the rotation of a ball in sports like football, baseball, and tennis.

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