# Torque - Dynamical Behavior

Just like force acting on a body, determines its linear dynamics of motion, similarly torque acting on a body determines the rotational dynamics. The best example to study the dynamical behaviour is rolling motion.

A rolling motion can be decomposed into translation of center of mass and rotation about center of mass.
Example

(i) Motion of a wheel of cycle on a road.
(ii) Motion of football rolling on a surface.

## Kinematics of Rolling motion

Let a round object is rolling on a horizontal fixed ground with velocity and acceleration of center equals 'v' and 'a' and angular velocity and angular acceleration are \(\omega \) and \(\alpha \) respectively. With this information velocity and acceleration of any other point can be calculated.

First, let us consider the velocity,

Rolling motion can be broken down into pure translation motion of the center of mass and pure rotational motion about the center of mass. When pure translation is considered then all the particles moves with same velocity 'v'. The particles on the rim that is point A, B and D move with same velocity 'v'.

When pure rotation is considered about center of mass, point B moves with velocity \(R\omega \) towards right, point D moves with velocity \(R\omega \) downwards, point A moves with velocity \(R\omega \) backwards.

For calculating the velocity in case of rolling motion, the velocities of the two cases are combined. Thus the velocity of point B is \(v+R\omega \) towards the right. Velocity of point D is a combination of two velocities, 'v' towards right and \(R\omega\) downwards, net velocity can be calculated by finding the resultant of these velocities.

The bottom-most point is the most interesting point. Its velocity is 'v' towards right and \(R\omega \) towards left. Based upon the resultant velocity of the bottom most point the rolling motion can be categorized as follows

1) Forward slipping : If \(v>R\omega \) then the bottom most point moves towards the motion of the object and it slips on the ground in forward direction thus the motion is said to be rolling with forward slipping. An example of forward slipping can be when hard brakes are applied to a fast moving vehicle. Brakes stops the rotational motion of the wheel, but translational velocity persists thus \(v>R\omega \)

2) Backward slipping: If \(v<R\omega \) then the bottom most point moves opposite to the motion of the object and it slips on the ground in backward direction thus the motion is said to be rolling with backward slipping. An example can be a tire stuck in mud, then on acceleration, tires rotate more and moves less in forward direction thus \(v<R\omega \)

3) Pure rolling or no slipping: If \(v=R\omega \) then the bottom most point comes to rest. If the points in contact are relatively at rest, then the object is said to be in pure rolling or no slipping.

For acceleration, the same method can be adopted, but while writing the acceleration in pure rotation it should be noted that the two components of acceleration must be written. One component is tangential to motion that is \(r\alpha \) where 'r' is the distance from center of mass and the second component is centripetal \(\omega^2 r \). Again for pure rolling with no slipping on fixed ground the condition comes out to be

\[a=R\alpha \]

Here \(\alpha \) equals angular acceleration

## Dynamics of Rolling Motion

Consider a round object of radius R and mass M on a rough horizontal road. Friction is enough to stop slipping. A constant force 'F' acts at the center in horizontal direction. The moment of inertia about the center of mass is \({I_{cm}} \). What is the acceleration of the center of mass of the body?

Let the friction 'f' acting on the body is in backward direction, the normal reaction 'N' is upward and gravitational force 'Mg' is downwards. Acceleration of center of mass is \({a_{cm}}\) in forward direction,

According to Newton's second law,

\[F - f = M{a_{cm}}\] Applying Newton's second law for rotation about the center of mass,

\[fR = {I_{cm}}\alpha \]

For pure rolling with no slipping,

\[{a_{cm}} = R\alpha \] \[\begin{array}{l} F - \frac{{{I_{cm}}{a_{cm}}}}{{{R^2}}} = M{a_{cm}}\\ {a_{cm}} = \frac{F}{{M + \frac{{{I_{cm}}}}{{{R^2}}}}} \end{array}\]

**Details and Assumptions**

The mass of the cotton reel is \(10 \text{ kg}\).

The moment of inertia of the reel about central axis is \(1000 \text{ kg m}^2\).

###### The problem is original.

**Cite as:**Torque - Dynamical Behavior.

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