College Physics I: BIIG problem-solving method

Rotational Motion

• The same physics describes the exhilarating spin of a skater and the wrenching force of a tornado.
• Force, energy, and power are associated with rotational motion.

Circular Motion

• Uniform Circular Motion is a motion in a circle at constant speed (angular velocity).

Angular Velocity

• Angular velocity is defined as the time rate of change of angle θ

                  ω   =   Δθ / Δt

where, θ is the angle of rotation.

• The relationship between angular velocity ω and linear velocity v was also defined is given by

            v  =  r ω           or              ω  =  v / r

where, r is the radius of curvature.

• According to the sign convention, the counter clockwise direction is considered as positive direction and clockwise direction as negative.

Angular Acceleration

• Angular acceleration is

                  α  =  Δω / Δt

where, Δω is the change in angular velocity and Δt   is the change in time.

• The units of angular acceleration are (rad/s)/s, or rad/s².
• If ω increases, then α is positive. If ω decreases, then α is negative.
• Problem (E10.1):  Suppose a teenager puts her bicycle on its back and spins the rear wheel at an angular velocity of 250 rpm.  If she now slams on the brakes, causing an angular acceleration of

– 87.3 rad/s², how long does it take the wheel to stop?                                                                  ( 0.30 s ) 

Tangential Acceleration

• For non-uniform circular motion, tangential acceleration at is directly related to the angular acceleration α and is linked to an increase or decrease in the velocity, but not its direction.

Linear Acceleration

• By definition,

            v  =  r ω           α  =  Δω / Δt

Therefore, the tangential acceleration is given by

            a_t  =  r α                       or             α  =  a_t  /  r

These equations mean that linear acceleration and angular acceleration are directly proportional.

• For example, the greater the angular acceleration of a car’s drive wheels,  the greater the acceleration of the car. The smaller a wheel, the smaller its linear acceleration for a given angular acceleration.
• Problem (E10.2):  A powerful motorcycle can accelerate from 0 to 30.0 m/s (about 108 km/h) in 4.20 s. What is the angular acceleration of its 0.320-m-radius wheels?                                            ( 22.3  rad/s² )

Kinematics of Rotational Motion

• The kinematics of rotational motion describes the relationships among rotation angle, angular velocity, angular acceleration, and time.
• Kinematics for rotational motion is completely analogous to translational kinematics.

θ  =  θ₀  +  ω t            ω  =  ω₀  +  α t             θ  =  ω₀ t  +  ½ α t²              ω²  =  ω₀² + 2 α θ  

• Problem (E10.5):  Large freight trains accelerate very slowly. Suppose one such train accelerates from rest, giving its 0.350-m-radius wheels an angular acceleration of 0.250 rad/s². After the wheels have made 200 revolutions (assume no slippage), what is the linear velocity of the train?             ( 9 m/s ) 

Dynamics of Rotational Motion: Rotational Inertia

• The torque, the turning effectiveness of a force is

             τ   =   r F    =    r ( m r α )   =   m r² α

• The quantity  m r²  is called the rotational inertia or moment of inertia of a point mass m a distance r from the center of rotation.

Moment of Inertia

• The moment of inertia I of an object is defined

            I  =  Σ m r²

the sum of mr² for all the point masses of which it is composed.

• I is analogous to m in translational motion.

I has units of mass multiplied by distance squared (kg m²).

• The general relationship among torque, moment of inertia, and angular acceleration is

            τ_net  =  I α                     α  =  τ_net / I

where, τ_net is the total torque from all forces relative to a chosen axis.

Newton’s Second Law for Rotation

• For simplicity,

             τ_net  =  I α                     α  =   τ / I

This equation is the rotational analog to Newton’s second law.

• It is actually valid for any torque, applied to any object, relative to any axis.
• Problem (E10.7):  Consider the father pushing a playground merry-go-round. He exerts a force of 250 N at the edge of the 50.0-kg merry-go-round, which has a 1.50 m radius. Calculate the angular acceleration produced (a) when no one is on the merry-go-round and (b) when an 18.0-kg child sits 1.25 m away from the center. Consider the merry-go-round itself to be a uniform disk with negligible retarding friction.                                                                                                                ( 6.7 rad/s² ;  4.4 rad/s² ) 

Rotational Kinetic Energy

• Rotational kinetic energy for an object with a moment of inertia I and an angular velocity ω

            K_rot  =  ½ I ω² 

• The expression for rotational kinetic energy is exactly analogous to translational kinetic energy,

with I being analogous to m and ω to v.

• Rotational kinetic energy has important effects.

Flywheels, for example, can be used to store large amounts of rotational kinetic energy in a vehicle.

• Problem (E10.9):  A typical small rescue helicopter, has four blades (use I = 4 M L²/3), each is 4.00 m long and has a mass of 50.0 kg rotating at 300 rpm. The helicopter has a total loaded mass of 1000 kg and it flies at 20.0 m/s. To what height could the helicopter be raised if all of the rotational kinetic energy could be used to lift it?                                                                                                       ( 50 m ) 

Angular Momentum

• Angular momentum is the rotational analog to linear momentum ( p  =  m v ).
• Angular momentum is defined as

            L  =  I ω

The units for angular momentum are kg m²/s.

• Large moment of inertia has a very large angular momentum.

Torque and Angular Momentum

• We know that the relationship between force and linear momentum is

            F  =  Δp / Δt

• The relationship between torque and angular momentum is

            τ  =  ΔL / Δt

This expression is the rotational form of Newton’s second law.

• The greater the torque, the more rapid the increase in angular momentum.
• Problem (E10.13):  The person kicks his leg by exerting a 2000-N force with his upper leg muscle. The effective perpendicular lever arm is 2.20 cm. Given the moment of inertia of the lower leg is 1.25 kg m², neglecting the gravitational force, what is the rotational kinetic energy of the leg after it has rotated through 57.3º (1.00 rad)?                                                                                                       ( 40 J ) 

Conservation of Angular Momentum

• If the net torque is zero, then angular momentum is constant or conserved.

            τ_net  =   ΔL / Δt   =   0

• If the change in angular momentum ΔL is zero, then the angular momentum is constant; thus,

            L  =  constant    ( τ_net = 0 )                                           L  =  L′     ( τ_net = 0 )

These expressions are the law of conservation of angular momentum.

Expressing this equation in terms of the moment of inertia,

            I ω  =   I′ ω′

where, the primed quantities refer to conditions after situation.

• Problem (E10.14):  Suppose an ice skater is spinning at 0.800 rev/ s with her arms extended. She has a moment of inertia of 2.34 kg m² with her arms extended and of 0.363 kg m² with her arms close to her body. What is her angular velocity in revolutions per second after she pulls in her arms?  (5.16 rev/s ) 

Gyroscopic Effects

• Angular momentum is a vector and, therefore, has direction as well as magnitude.
• Torque affects both the direction and the magnitude of angular momentum.
• The direction of angular velocity ω size and angular momentum L are defined to be the direction in which the thumb of your right hand points when you curl your fingers in the direction of the disk’s rotation.

BIIG: Problems & Solutions