Chapter 11 Rotational Dynamics As we did for linear or translational motion we studied kinematics motion without regard to the cause and then dynamics motion with regard to the cause we now proceed in a similar fashion We know that forces are responsible for linear motion We will now see that rotational motion is caused by torques Consider a wrench of length L r F Torque rF Units of N m r Not work or energy do not use Joules r is called the lever arm Must always be perpendicular to the force If the force is not perpendicular to the level arm we need to find the component that is perpendicular either the force or the lever arm F F sin rF sin If 0 then the torque is zero r Therefore torques force times length are responsible for rotational motion Newton s 2nd Law for Rotational Motion The torque for a pointparticle of mass m a distance r from the rotation axis is rFt Ft mat at r 2 mat r mr F Ft r m Axis of rotation Define I mr2 Moment of Inertia for a point particle a scalar units of kg m2 A rigid body is composed of many many particles of mass mi which are ri from the axis of rotation Each of these masses creates a torque about the axis of rotation Rotation axis 2 1 1 2 2 2 1 m r 2 m r r1 r2 m1 m2 Sum up all torques due to all particles 2 m r is the same for all particles i I mi ri I i i 2 I moment of inertia for the rigid body It is different for different shaped objects and for different axes of rotation Table 10 1 For a thin rod of mass M and length L I L 2 I 1 12 ML2 1 3 2 ML L The last equation is Newton s 2nd Law for rotation Compare to the translational form I Example Problem F ma A rotating door is made of 4 rectangular panes each with a mass of 85 kg A person pushes on the outer edge of one pane with a force of 68 N directed perpendicular to the pane Determine the door s Given L 1 2 m mpane 85 kg F 68 N L I FL FL I I F From Table 10 1 moment of inertia for a thin rectangular rod is same as a thin sheet 2 2 I pane ML I door ML Since there are 4 panes FL FL 4 2 I 3 ML 3F 3 68 N 0 50 kgNm 0 50 rad s2 4 ML 4 85 kg 1 2 m 1 3 4 3 Example Problem The parallel axis theorem provides a useful way to calculate I about an arbitrary axis The theorem states that I Icm MD2 where Icm is the moment of inertia of an object of mass M with an axis that passes through the center of mass and is parallel to the axis of interest D is the perpendicular distance between the two axes Now determine I of a solid cylinder of radius R for an axis that lies on the surface of the cylinder and perpendicular to the circular ends Solution The center of mass of the cylinder is on a line defining the axis of the cylinder R From Table 10 1 cm I cm MR 1 2 2 From the parallel axis theorem with D R 2 2 2 I I cm MD MR MR MR 1 2 Apply to thin rod L 2 I cm 1 12 ML2 2 1 12 2 I I cm MD 3 2 2 L 2 L 2 1 ML M 3 ML 2 Rotational Work For translational motion we defined the work as W Fd d For rotational motion Fd d r r s arc length s Ft W Fd d Ft r Ft r r W R Units of N m or J when is in radians Rotational Kinetic Energy For translational motion the K was defined 2 K mv since v t r 2 2 2 1 K m r 2 mr 1 2 1 2 For a point particle I mr2 therefore 1 2 K rot I 2 Or for a rigid body K rot 1 2 m r 2 i i 2 2 1 2 I K rot For a rigid body that has both translational and rotational motion its total kinetic energy is 2 1 2 1 2 K total K K rot mv I 2 The total mechanical energy is then 1 2 2 1 2 2 1 2 E total mv I mgy kx 2 Example A car is moving with a speed of 27 0 m s Each wheel has a radius of 0 300 m and a moment of inertia of 0 850 kg m2 The car has a total mass including the wheels of 1 20x103 kg Find a the translational K of the entire car b the total Krot of the four wheels and c the total K of the car Solution Given vcar 27 0 m s mcar 1 20x103 kg rw 0 300 m Iw 0 850 kg m2 a b c 1 2 2 car 5 1 2 3 2 K mcar v 1 2x10 27 0 K 4 37x10 J 2 2 v v 2 t car 1 1 1 K rot 2 I 2 Iw 2 Iw rw rw 2 1 K rot 2 0 850 27 0 0 300 3 3 44x10 J 4 K wheels 4K rot 1 38x10 J K total K K wheels 5 4 5 4 37x10 1 38x10 4 51x10 J Example Problem A tennis ball starting from rest rolls down a hill into a valley At the top of the valley the ball becomes airborne leaving at an angle of 35 with respect to the horizontal Treat the ball as a thinwalled spherical shell and determine the horizontal distance the ball travels after becoming airborne 0 3 1 8 m 1 2 4 x Solution Given v0 0 0 0 y0 1 8 m h y1 y2 y4 0 2 35 x2 0 I 2 3 MR2 Find x4 Method As there is no friction or air resistance in the problem therefore no non conservative forces we can use conservation of mechanical 2 2 1 1 energy E mv I mgy total 2 2 E0 mgh 2 2 1 1 E1 2 mv1 2 I 1 mg 0 2 2 1 1 E2 2 mv 2 2 I 2 mg 0 E1 v1 v 2 1 2 Since E0 E1 E2 2 2 2 2 mgh mv I but v t r 1 2 vt v 2 1 2 Velocity of ball equals tangential velocity at edge of ball 2 2 2 mgh mv mr v 2 r 1 2 1 2 2 3 2 gh 12 v 22 13 v 22 56 v 22 v 2 6 gh 5 6 9 80 1 8 5 4 6 m s Now use 2D kinematic equations for projectile motion v …
View Full Document
Unlocking...