Understanding Rolling Without Slipping

When a solid disc of radius \(r\) rolls without slipping on a horizontal floor, it means there is no relative motion between the point of the disc in contact with the ground and the ground itself. This condition has important implications for both the translational and rotational motion of the disc.

• Translational Motion: The center of mass (CM) of the disc moves linearly along the horizontal floor. Let its velocity be \(v_c\) and its acceleration be \(a_c\).
• Rotational Motion: The disc rotates about its center of mass with an angular velocity \(\omega\) and an angular acceleration \(\alpha\).

Condition for Rolling Without Slipping

For a disc rolling without slipping, the linear velocity of the center of mass \(v_c\) is directly related to its angular velocity \(\omega\):

\[v_c = r\omega\]

Similarly, the linear acceleration of the center of mass \(a_c\) is related to its angular acceleration \(\alpha\):

\[a_c = r\alpha\]

These relationships are fundamental to analyzing the motion of rolling objects.

Acceleration of Point of Contact

We need to find the magnitude of acceleration of the point of contact on the disc. Let's denote the point of contact at the bottom of the disc as P. The acceleration of any point P on a rigid body undergoing both translation and rotation can be found using the following vector sum:

\[\vec{a}_P = \vec{a}_c + \vec{a}_{P/c}\]

Where:

• \(\vec{a}_P\) is the absolute acceleration of the point of contact P.
• \(\vec{a}_c\) is the acceleration of the center of mass (CM) of the disc.
• \(\vec{a}_{P/c}\) is the acceleration of point P relative to the center of mass.

Components of Acceleration Relative to Center of Mass (\(\vec{a}_{P/c}\))

The acceleration of point P relative to the center of mass C, \(\vec{a}_{P/c}\), has two distinct components for circular motion:

• Tangential Acceleration (\(a_t\)): This component arises from the angular acceleration \(\alpha\). Its magnitude is \(r\alpha\). For the point of contact P at the bottom, this tangential acceleration is directed horizontally, opposite to the direction of the center of mass acceleration \(a_c\). If the disc is rolling to the right, \(a_c\) is to the right, and this \(a_t\) is to the left.
• Centripetal Acceleration (\(a_r\)): This component is due to the angular velocity \(\omega\) and is always directed towards the center of rotation, which is the CM in this case. Its magnitude is \(r\omega^2\). For the point of contact P at the bottom, this centripetal acceleration is directed vertically upwards, towards the center of the disc.

Vector Sum for Total Acceleration

Let's choose a coordinate system where the horizontal direction is the x-axis and the vertical direction is the y-axis. Assume the disc rolls to the right.

• The acceleration of the center of mass, \(\vec{a}_c\), is purely horizontal (in the +x direction): \[\vec{a}_c = r\alpha \hat{i}\]
• The tangential acceleration of the point of contact P relative to the CM, \(\vec{a}_{P/c, \text{tangential}}\), is also horizontal but in the opposite direction (in the -x direction): \[\vec{a}_{P/c, \text{tangential}} = -r\alpha \hat{i}\]
• The centripetal acceleration of the point of contact P relative to the CM, \(\vec{a}_{P/c, \text{centripetal}}\), is purely vertical (in the +y direction, towards the center): \[\vec{a}_{P/c, \text{centripetal}} = r\omega^2 \hat{j}\]

Now, summing these vector components to find the total acceleration of the point of contact \(\vec{a}_P\):

\[\vec{a}_P = \vec{a}_c + \vec{a}_{P/c, \text{tangential}} + \vec{a}_{P/c, \text{centripetal}}\] \[\vec{a}_P = (r\alpha \hat{i}) + (-r\alpha \hat{i}) + (r\omega^2 \hat{j})\] \[\vec{a}_P = (r\alpha - r\alpha) \hat{i} + r\omega^2 \hat{j}\] \[\vec{a}_P = 0 \hat{i} + r\omega^2 \hat{j}\] \[\vec{a}_P = r\omega^2 \hat{j}\]

As seen above, the horizontal components of acceleration cancel each other out due to the condition of rolling without slipping (\(a_c = r\alpha\)).

Magnitude of Acceleration

The total acceleration of the point of contact P is solely in the vertical direction. Therefore, the magnitude of this acceleration is:

\[|\vec{a}_P| = \sqrt{0^2 + (r\omega^2)^2}\] \[|\vec{a}_P| = \sqrt{(r\omega^2)^2}\] \[|\vec{a}_P| = r\omega^2\]

This acceleration is directed vertically upwards, towards the center of the disc.

Thus, the magnitude of acceleration of the point of contact on the disc is \(r\omega^2\).