A Tale of Friction Basic Rollercoaster Physics Fahrenheit - PowerPoint PPT Presentation

A Tale of Friction Basic Rollercoaster Physics Fahrenheit Rollercoaster, Hershey, PA | max height = 121 ft | max speed = 58 mph

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Rotational Movement Kinematics Similar to how linear velocity is defined, angular velocity is the angle swept by unit of time. Tangential velocity is the equivalent of linear velocity for a particle moving on a circumference.    s r  d  r    v T v T r or dt d  dv T    T a dt dt  d  2 2 d    a T r 2 2 dt dt    a T r 

Rotational Kinetic Energy and Momentum of Inertia of a Rigid Body • For a single particle: K  2 Tangential kinetic energy: mv 1 2 T K   2 I Rotational kinetic energy: 1 2 I  2 mr Momentum of inertia: • For a system of particles:   Momentum of inertia: 2 I mr i • For a rigid body: I   2 Momentum of inertia: r dm

Angular Momentum and Torque of a Rigid Body   F  • Law of lever: d       • r F Torque :      r F sin Magnitude :     • Newton’s F m a second law:  d v   m dt Torque is a measure of how much a force acting on an object causes that object to rotate. It is formally defined as a vector coming from the special product of the position vector of the point of application of the force, and the force vector . Its magnitude depends on the angle between position and force vectors. If these vectors are parallel, the torque is zero.

Angular Momentum and Torque of a Rigid Body     • Linear momentum: P m v   d   •   Force definition: F m v dt  d v   m For m = constant: dt m    a      • Angular momentum: L r p      m r v       L m r v If r F : T Defining torque (force producing rotation) in a circular movement dL dv ( r constant) as the change in time        T m r m r a T dt dt of the angular moment:

Angular Momentum and Torque of a Rigid Body     • Linear momentum: P m v     d •   Force definition: F m v dt  d v   m For m = constant: dt m    a      • Angular momentum: L r p      m r v       L m r v : If r F T Taking a T =   r , and making I = m  r 2 :         2 m r a m r T or   I  

Friction Force for a Rigid Sphere Rolling on an Incline The sphere rolls because of the torque produced by the friction force f s and the weight’s component parallel to the incline:              and F m a m g sin f f s r I s If the sphere’s momentum of inertia is I = 2/5  m  r 2 and  = a / r : 2 2 a  5      or f s m a 2 f s r m r 5 r With this value: 2     sin    m a m g m a 5 Solving for a in the above equation, the acceleration of the sphere rolling on the incline is: 5     a g sin 7

Friction Force for a Rigid Sphere Rolling on an Incline 2 5  5      Combining: and f s m a a g sin 7 the static friction force is now: 2     f s m g sin 7 But by definition, the static friction force is proportional to the normal force the body exerts on the surface :    f F s s n Taking F n from the free-body diagram:       f m g cos s s

Friction Force for a Rigid Sphere Rolling on an Incline Combining the two expressions for f s : 2          m g sin m g sin s 7 the coefficient of static friction can be expressed as: 2    tan s 7 This expression states that the coefficient of static friction is a function of the incline’s angle only, specifically, a function of the slope of this surface.

Friction Force for a Rigid Sphere Rolling on a Variable Slope Path Let f ( x ) a differentiable function. If: dy     m f ' ( x ) and m tan dx then : tan   f ’( x ) The coefficient of static friction  s can At any point of a curved path f ( x ) , a tangent line be expressed as: can be visualized as a portion of an incline. 2  s  f ' ( x ) The slope m of this incline is the tangent of the 7 angle between this line and the horizontal, tan  . The static friction force f s is now: In calculus, this slope is given by the value of f’ ( x ) , the derivative of the function f ( x ) at that 2      f s m g f ' ( x ) cos point. 7

Friction Force for a Rigid Sphere Rolling on a Variable Slope Path Because tan  = f ’( x ), it is possible to define a f ' ( x ) opposite tan    right triangle with sides in terms of f ’( x ) : 1 adjacent   If: , then: arctan( f ' ( x )) 2     f s m g f ' ( x ) cos(arctan ( f ' ( x ))) 7 Using basic trigonometry: 1 adjacent    cos  hypotenuse 2 1 ( f ' ( x )) 2 f ' ( x )    f s m g The static friction force is now:  7 2 1 ( f ' ( x ))

Friction Force for a Rigid Sphere Rolling on a Variable Slope Path But, something needs to be fixed in this procedure. By definition, the static friction coefficient  s must always be positive, while the slope of a path may be positive or negative. So the required corrections must be: 2  s  f ' ( x ) 7 f ' ( x ) 2    f s m g  7 2 1 ( f ' ( x )) Where: denotes the absolute value of the function f ’( x ) f ' x ( )

Work-Energy for a Sphere Rolling on a Variable Slope Path with Friction The work-energy theorem states that the mechanical energy (kinetic energy + potential energy) of an isolated system under only conservative forces remains constant:      E K U K U E f f f i i i or       E K U 0 In a system under non-conservative forces, like friction, the work-energy theorem states that work done by these forces is equivalent to the change in the mechanical energy:        W f E K U Additionally, the work done by non-conservative forces depends on the path or trajectory of the system, or in the time these forces affect the system.

Friction Force for a Rigid Sphere Rolling on a Variable Slope Path By definition, mechanical work is the product of the displacement and the force component along the displacement: For a variable slope path y = f ( x ) , the work done by the friction f s over a portion  s of the path is:     W f s s f ' ( x ) 2      m g s  7 2 1 ( f ' ( x )) For a differential portion of the path: f ' ( x ) 2     dW m g ds  7 2 1 ( f ' ( x ))

Friction Force for a Rigid Sphere Rolling on a Variable Slope Path Expressing ds in terms of the differentials dx and dy , the differential arc can be expressed in terms of the f ’ ( x ) :   2           dy          2 2   2 2 ds dx dy 1 dx 1 f ' ( x ) dx     dx   The work along the differential portion of the path can be expressed as: f ' ( x ) 2     dW m g ds  7 2 1 ( f ' ( x )) f ' ( x ) 2       2 1 ( ' ( )) m g f x dx  7 2 1 ( f ' ( x )) 2     dW m g f ' ( x ) dx 7

Friction Force for a Rigid Sphere Rolling on a Variable Slope Path Because dx > 0 , using properties of 2     dW m g f ' ( x ) dx the absolute value and the definition 7 of differential of a function: 2     m g f ' ( x ) dx 7 2    m g df ( x ) 7 Friction forces always acts against the movement, so the work done by 2     dW m g df ( x ) them must always be negative: 7

Friction Force for a Rigid Sphere Rolling on a Variable Slope Path Taking small displacements instead differentials: 2            W m g f ( x ) W f K U 7 Using this expression in the work-energy theorem: 2               2 2 m g f ( x ) m v m v m g h m g h 1 1 2 2 f i f i 7 This expression relates the work done by friction with the mechanical energy of a sphere rolling on a little portion of a curved path . Visualize this portion as a little incline . Height h is given by the function f ( x ) .

Friction Force for a Rigid Sphere Rolling on a Variable Slope Path Then, dividing by m : 2             2 2 ( ) ( ) ( ) ( ) g f x f x 1 v 1 v g f x g f x 2 2 f i f i f i 7 From this expression, we can determine final velocity at the end of the incline:   4          2 v 1 v 2 g f ( x ) f ( x ) g f ( x ) f ( x ) 2 f i f i f i 7