In pure rolling motion, more specifically rolling motion of a body, how do we decide the direction of acceleration of centre of mass and angular acceleration of the body?

For examples in the question,

A string wrapped around a cylinder of mass m and Radius R. The end of the string is connected to a block of same mass hanging vetrically through a massless and frictionless pulley. No friction exists between the horizontal surface on which the cylinder is placed on and the cylinder. Acceleration of the hanging mass is how much.

Please help ASAP!!

To determine the direction of acceleration of the center of mass and the angular acceleration of a body in pure rolling motion, we need to analyze the forces and torques acting on the system. In your example, we have a cylinder and a hanging block connected by a string. Let's break this down step by step.

Understanding the System

We have a cylinder of mass m and radius R resting on a frictionless surface. A string is wrapped around the cylinder, and the other end of the string is attached to a block of the same mass m hanging vertically. The pulley is massless and frictionless, which simplifies our calculations.

Identifying Forces

• The hanging block experiences gravitational force downward, which is mg.
• The tension in the string acts upward on the block and downward on the cylinder.

Acceleration of the Hanging Mass

Let’s denote the acceleration of the hanging mass as a. According to Newton's second law, we can write the equation for the hanging mass:

mg - T = ma

Here, T is the tension in the string. Rearranging gives us:

T = mg - ma

Angular Acceleration of the Cylinder

For the cylinder, the tension in the string also creates a torque about its center of mass. The torque τ can be expressed as:

τ = T * R

Using the relationship between torque and angular acceleration (α), we have:

τ = I * α

For a solid cylinder, the moment of inertia I is given by:

I = (1/2)mR²

Substituting the expressions for torque and moment of inertia, we get:

T * R = (1/2)mR² * α

From this, we can express the angular acceleration:

α = (2T)/(mR)

Relating Linear and Angular Acceleration

In pure rolling motion, the linear acceleration a of the center of mass is related to the angular acceleration α by the equation:

a = R * α

Substituting our expression for α gives:

a = R * (2T)/(mR) = (2T)/m

Combining the Equations

Now we have two equations involving tension T:

1. T = mg - ma
2. a = (2T)/m

Substituting the first equation into the second, we can solve for a:

From the first equation, we can express T as:

T = mg - ma

Substituting this into the second equation:

a = (2(mg - ma))/m

Now simplifying:

a = 2g - 2a

Bringing all terms involving a to one side gives:

3a = 2g

Thus, we find:

a = (2/3)g

Final Thoughts

The acceleration of the hanging mass is therefore (2/3)g, which indicates that the center of mass of the system accelerates downward at this rate. The angular acceleration of the cylinder can then be found using the relationship α = a/R.

This analysis illustrates how the forces and torques interact in a system involving rolling motion and helps us understand the dynamics at play. If you have any further questions or need clarification on any part, feel free to ask!