Question

A small metal cylinder rests on a circular turntable that isrotating at a constant rate, as illustrated in the diagram.

$12971_a.jpg$

Which of the following sets of vectors best describes thevelocity, acceleration, and net force acting on the cylinder at thepoint indicated in the diagram?

$12971_b.jpg$

Let be the distance between the cylinder and the center of theturntable. Now assume that the cylinder is moved to a new location from the center of the turntable. Which of the followingstatements accurately describe the motion of the cylinder at thenew location?

1. The speed of the cylinder has decreased.
2. The speed of the cylinder has increased.
3. The magnitude of the acceleration of the cylinder hasdecreased.
4. The magnitude of the acceleration of the cylinder hasincreased.
5. The speed and the acceleration of the cylinder havenot changed.

Answer 1

Concepts and reason

The concepts of centripetal force, centripetal acceleration, and the tangential speed are to be used.

A point is located on the rotating turntable. Determine the direction of speed at that point. For a circular motion at constant rate, the direction of speed is tangential to the circular path. Determine the direction of force which is acting radially inward toward the center of the circular path. Determine the direction of acceleration by using the direction of force. In circular motion at constant speed, the direction of acceleration is same as the direction of force.

Finally, determine the change in the magnitude of the speed and the acceleration of the cylinder with change in the distance between the cylinder and the center of the table.

Fundamentals

Acceleration is defined as the time rate of change of speed. For a circular motion at constant rate, the direction of speed is always tangential to the circular path. The force acting on an object moving in circular motion is a centripetal force which is directed inward toward the center of the circular path.

For a circular motion at constant speed, the direction of acceleration is same as the direction of the net force acting on the object moving in circular motion.

The speed of an object moving in circular motion is,

$v = \frac{{2\pi R}}{T}$

Here, R represents the radius of the circle and T represents the period of the motion.

The acceleration of an object moving in circular motion is,

$a = \frac{{{v^2}}}{R}$

Here, v is the speed of the object and R is the radius of the circle.

Determine the direction of net force acting on the cylinder at the point.

The point is located on the rotating turntable. The point on the rotating turn table experiences the centripetal force. As the direction of this force in circular motion is radially inward, the direction of net force acting at that point is inward toward i.e. the center of the circular path.

Hence, at the point, the direction of force is downward i.e. toward the center of the circular path.

Determine the direction of acceleration.

For the rotating cylinder, the direction of acceleration is same as the direction of net force acting on the cylinder.

As the net force is acting in the downward direction, the direction of acceleration is downward.

Determine the direction of speed.

The cylinder is rotating in the right direction with constant speed. As the motion of the cylinder is circular, the direction of speed at the point is tangential to the circular path.

Hence, the speed is perpendicular to the direction of the net force and is directed towards right.

The speed of an object moving in circular motion is,

$v = \frac{{2\pi R}}{T}$

Here, R represents the distance between the cylinder and the center of the table and T represents the period of the motion.

Substitute $v'$for v and $\frac{R}{2}$ for R.

$\begin{array}{c}\\v' = \frac{{2\pi \left( {\frac{R}{2}} \right)}}{T}\\\\ = \frac{1}{2}\left( {\frac{{2\pi R}}{T}} \right)\\\end{array}$

Here, $v'$is the changed speed.

Substitute v for $\frac{{2\pi R}}{T}$.

$v' = \frac{v}{2}$

Hence, the speed is decreased by half.

Determine the change in the acceleration of the cylinder if the distance between the cylinder and the center of the table is reduced by half.

The acceleration of an object moving in circular motion is,

$a' = {\left( {\frac{{2\pi }}{T}} \right)^2}R'$

Here, R is the radius of the circular path and T is the period of the circular motion.

Substitute $a'$for a and $\frac{R}{2}$ for$R'$.

$a' = {\left( {\frac{{2\pi }}{T}} \right)^2}\frac{R}{2}$

Here, $a'$ is the changed acceleration.

Substitute a for ${\left( {\frac{{2\pi }}{T}} \right)^2}R$.

$a' = \frac{a}{2}$

Hence, the magnitude of acceleration is decreased by a factor$\frac{1}{2}$.

Ans:

The direction of acceleration and the net force acting on the cylinder at the point is downward and the speed is directed toward right.