A very small round ball is located near a large solid sphere of uniform density. The force that the large sphere exerts on the b

We will use the equations of rotational kinematics,

            (A)

                                    (B)                                          

Here, and denote the final and initial angular displacements, respectively, whereas and represent final and initial angular velocities, and is the angular acceleration.

We are provided with and .

By substituting these values into equation (A), we have

Now, using equation (B),

This indicates that the wheel's angular speed at the 4.20-second mark is 36.7 rad/s.

Answer:

(A) The tension's magnitude grows to four times the initial value, 4F.

Explanation:

When an object travels in a circular path, a centripetal force is exerted upon it. In this instance, the centripetal force acting on the stone can be represented by .

                   Here, m denotes the mass of the object

                               v is the velocity or speed of the object

                               r signifies the radius of the circular path

Importantly, the tension corresponds to the centripetal force.

Initially, the string completes one revolution each second, and subsequently, it accelerates to perform two revolutions in the same time frame. This signifies that the speed has increased twofold.

Applying our formula:F =

                               where F indicates the tension in the string

assuming the starting speed is v, after doubling it becomes 2v

Maintaining the circle's radius, we arrive at:

F=

From this equation, it's clear that the initial tension has quadrupled.

Consequently, the magnitude of the tension increases to four times its original value, 4F.