Big Picture

Rotating objects act differently than those in linear or projectile motion. All objects moving in a circular path have linear acceleration because they are constantly changing direction. All of Newton’s laws of motion, which normally only apply to linear motion, have rotational equivalents.

Key Terms

Centripetal Force

Centripetal forces are responsible for circular motion. The force vector will always be pointed toward the center of the circle and be perpendicular to the direction of the object’s motion.

There are a few forces (called fictitious forces) that are associated with circular motion but do not actually exist when observing the rotating object from a stationary reference frame.

• Centrifugal force: An outward force that is wrongly believed to pull objects away from the center. When whirling a tennis ball on a string, many people believe that if the string were to break, the ball would fly away from the circle.
  

In reality, it would go off in a tangential straight line because there is no force acting on it! The force felt is actually feeling the effects of Newton’s third law. Since the person whirling the tennis ball exerts a centripetal force on the ball pointed towards the person, the object is also exerting an equal and opposite force on the person’s hand.

Linear vs. Circular Motion

Image Credit: Wizard191, Public Domain

In circular motion, an object rotates in a circle around the rotational axis.

• To find the rotational axis, use the right-hand rule: use the fingers of your right hand to follow the direction of rotation, and your thumb will point along the axis.
• The distance between the rotational axis and the object is called the radius (r).
• To find the path length (distance an object travels along a circular path), use the geometric formula for finding arc length: s = rθ, where s is the path length.
• The period T is the time it takes to complete one rotation. The frequency f is equal to 1/T.

Linear vs. Circular Motion (cont.)

Circular motion is similar to linear motion in many ways.

• Angular velocity and torque have two directions: clockwise and counterclockwise.
• Only the component of torque perpendicular to the radius of rotation will cause an object to accelerate.
• Angular acceleration is not the same as centripetal acceleration! Centripetal acceleration always points toward the rotation axis. Angular acceleration points in or against the direction of angular velocity.
• Angular velocity is related to period and frequency: \(\omega = \frac{2\pi}{T} = 2\pi f\)

Using angular quantities, we can rewrite the equations for linear motion.

• \(KE_{rot} = \frac{1}{2}Iw^2\)
• For constant a,
• \(\Delta \Theta = \Sigma_o t + \frac{1}{2}\alpha t^2\)
• \(\Sigma = \Sigma_o t + \Delta t\)
• \(\Sigma_2 = \Sigma_o^2 + 2 \Sigma(\Delta \Theta)\)

Precession

Image Credit: Xavier Snelgrove, Christopher Addiego, CC-BY-SA 2.5

Precession is most commonly viewed in the motion of a spinning top. When the top’s center of gravity is not directly over the bottom tip of the top, gravity exerts a torque on the spinning top. Since the top’s axis itself initially has zero angular momentum, the whole axis will rotate to counter the torque from gravity.

Notes

Important Equations

To convert between degrees and radians, use the conversion factor: \(1 rad = \frac{180^\circ}{\pi}\)

Example Problem

A mass (\(m_1\)) is spinning horizontally on a frictionless table secured by a string of negligible mass. The string goes through a hole in the center of the table and is attached to another hanging object of unknown mass (we'll call it m2). The radius of m1's rotation is r, and m1 is rotating with speed v. Find the mass of the object hanging under the table.

Solution

In this problem, the tension in the string is acting as a centripetal force to make the mass on top of the table rotate. Since the string is also attached to the hanging mass, the tension is also equal to the weight of the unknown mass. We can use these two pieces of information to find the mass of the hanging object.

Mass on a String

At the bottom of its path, the free body diagram for a mass on a string is as shown below:

At the top of its path, the free body diagram for the mass is as shown below:

At the minimum speed needed for the mass to make it over, at the top of the path, the tension in the string will be 0.

Rolling Bodies

If an object rolls down a ramp, there must be friction. Without friction, the object would slip down the ramp without rotating. Because friction causes the object to roll, friction is the force we plug in when we use the torque equation.

Example

A hoop of mass M, radius R, and moment of inertia of \(MR^2\) is released from the top of a ramp of angle θ. What is the acceleration of the center of mass of the hoop? Assume there is no slipping.

Solution

Given: \(m = M, r = R, I = MR^2\), let μ be the coefficient of static friction

Find: a

\(T = I \alpha = MR^2(\frac{a}{R}) \text{and} T = F \perp r = F_fR\)

friction is the force providing torque

\(MR^2(\frac{a}{R} = F_fR\)

set the two expressions for torque equal to each other

\(F_f = \mu N = \mu Mg \cos \theta\)

\(F_{net} = Ma = Mg \sin \theta = \mu Mg \cos \theta\)

\(a = g \sin \theta - \mu Mg \cos \theta\)

Pulleys

In a pulley system, tension causes the pulley to rotate, which means tension is the force we plug in for torque. If there are two blocks attached to a single string, we know that they will have the same acceleration because they are in the same system. Also, if they are of different mass, they will have different free body diagrams, so draw one for each!

Since block 1 is 6 g and block 2 is 3 g, we can conclude that block 1 will fall in this system while block 2 rises.

\(T = I \alpha = I\frac{a}{R} \text{and} T = F \perp R\)
\(F \perp = T_1 - T_2\)
\(I\frac{a}{R} = (T_1 - T_2)R\)
\(I = \frac{(T_1 - T_2)R^2}{a}\)

Notes