Question

A solid disc of radius r rolls without slipping on a horizontal floor with angular velocity and angular acceleration . The magnitude of the acceleration of the point of contact on the disc is

Answer 1

it's net acceleration along point of contact is zero

Answer 2

Answer: The correct answer is $a_{r} = r\omega ^{2}$.

Explanation:

In the given problem, a solid disc of radius r rolls without slipping on a horizontal floor with angular velocity and angular acceleration .

Rolling is a type of motion which combines translation and rotation of the object.

There are two types of accelerations which will act during the rolling of the object: angular acceleration (2) radial acceleration.

The expression of the radial acceleration is as follows;

$a_{r} = r\omega ^{2}$

Here, r is the radius and $\omega$ is the angular velocity.

At the point of contact of the solid disc, there is only radial acceleration.

Therefore, the magnitude of the acceleration of the point of contact on the disc is  $r\omega ^{2}$.