Rolling Motion

You must have seen the motion of a rolling ball or a wheel many times, but do you know the kind of motions that an object and its particles undergo while in rolling motion? A combination of translational and rotational motions happen during the rolling motion of a rigid object. To define rolling motion, we must understand the forces like angular momentum and torque. This article will give you the definition of rolling motion, and you would also learn rolling motion equations here.

Rolling Objects Physics

When there is a rolling motion without slipping,  the object has both rotational and translational movement while the point of contact is instantaneously at rest.

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Let us first understand pure translational and pure rotational motions.

• Pure Translational Motion

An object in pure translational motion has all its points moving with the same velocity as its center of mass i.e. they all have the same speed and direction or V(r) = Vcenter of mass. In the absence of an external force, the object would move in a straight line.

• Pure Rotational Motion

An object in pure rolling motion has all its points moving at right angles to the radius (in a plane that is perpendicular to its rotational axis). The speed of these particles is directly proportional to their distance from the axis of rotation. Here V(r) = r * ω. Here ω is the angular frequency. Since at the axis r is 0 hence particles on the axis of rotation do not move at all whereas points at the outer edge move with the highest speed.

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• The points on either side of the axis of rotation move in opposite directions.

• Vpoint of contact = 0 i.e. the point of contact is at rest.

• The velocity of the center of mass is Vcenter of mass = R * ω.

• The point farthest from the point of contact move with a velocity of

Vopposite the point of contact = 2 * Vcenter of mass = 2 * R * ω.

Heave and Pitch

A ship on the sea has 6 different kinds of motions called:

• Heave

This is a linear motion along the vertical z-axis.

• Sway

This is also a linear motion along the transverse Y-axis.

• Surge

It is again a linear motion along the longitudinal x-axis.

• Roll

This is a rotational motion around a longitudinal axis.

• Pitch

This is a rotational motion around the transverse axis.

• Yaw

This is a rotational motion around the vertical axis.

Mechanical Energy is Conserved in Rolling Motion

As per the rolling motion definition, a rolling object has rotational kinetic energy and translational kinetic energy. If the system requires it, it might also carry potential energy. If we include the gravitational potential energy also then we get the total mechanical energy of a rolling object as:

Etotal = (½ * m * V2center of mass) + (½ * Icenter of mass * ω2) + (m * g * h).

When there are no nonconservative forces that could take away the energy from the system in the form of heat, an object's total energy in rolling motion without slipping is constant throughout the motion. When the object is slipping them energy is not conserved since there is a heat production due to kinetic friction and air resistance.

Moment of Inertia

Rotational inertia is a property of rotating objects. It is the tendency of an object to remain in rotational motion unless a torque is applied to it.

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If a force F is exerted on a point mass m at a distance r from the pivot point, then the point mass obtains an acceleration equal to F/m in the direction of F. Since F is perpendicular to r in the case above, the torque τ = F * r. The rotational inertia is given by the formula m * r².

Parallel Axis Theorem

If the rotational axis passes through the center of mass, then the moment of inertia is minimal. Moment of inertia increases as the distance of the axis of rotation from the center of mass increases. As per the parallel axis theorem, the moment of inertia about an axis that is parallel to the axis across the object’s center of mass is given by the below formula:

Iparallel axis = Icenter of mass + M * d2

Where d is the distance of the parallel axis of rotation from the center of mass.

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Let us understand this with an example: Let there be a uniform rod of length l having mass m, rotating about an axis through its center and perpendicular to the rod. What is the moment of inertia Icenter of mass?

Solution. Moment of inertia of a rod = ⅓ * m * l2

Distance of the end of the rod from its center = l/2

Hence the parallel axis theorem of the rod = ( ⅓ * m * l2) - m * (l/2)2)

                = ( ⅓ * m * l2) - ( ¼  * m * l2) 

Icenter of mass = 1/12 * m * l2