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Rotational Kinematics and Energy

Rotational Kinematics and Energy. Rotational Motion. Up until now we have been looking at the kinematics and dynamics of translational motion – that is, motion without rotation. Now we will widen our view of the natural world to include objects that both rotate and translate.

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Rotational Kinematics and Energy

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  1. Rotational Kinematics and Energy

  2. Rotational Motion Up until now we have been looking at the kinematics and dynamics of translational motion – that is, motion without rotation. Now we will widen our view of the natural world to include objects that both rotate and translate. We will develop descriptions (equations) that describe rotational motion Now we can look at motion of bicycle wheels, roundabouts and divers.

  3. Rotational variables • - Angular position, displacement, velocity, acceleration II.Rotation with constant angular acceleration III. Relation between linear and angular variables - Position, speed, acceleration IV.Kinetic energy of rotation V. Rotational inertia VI. Torque VII. Newton’s second law for rotation VIII. Work and rotational kinetic energy

  4. Rotational kinematics • In the kinematics of rotation we encounter new kinematic quantities • Angular displacement q • Angular speed w • Angular acceleration a • Rotational Inertia I • Torque t • All these quantities are defined relative to an axis of rotation

  5. Angular displacement • Measured in radians or degrees • There is no dimension Dq = qf - qi qi Dq Axis of rotation qf

  6. qi P r Q sp sq Axis of rotation qf Angular displacement and arc length • Arc length depends on the distance it is measured away from the axis of rotation

  7. Angular Speed • Angular speed is the rate of change of angular position • We can also define the instantaneous angular speed

  8. Average angular velocity and tangential speed • Recall that speed is distance divided by time elapsed • Tangential speed is arc length divided by time elapsed • And because we can write

  9. Average Angular Acceleration • Rate of change of angular velocity • Instantaneous angular acceleration

  10. Angular acceleration and tangential acceleration • We can find a link between tangential acceleration at and angular acceleration α • So

  11. Centripetal acceleration • We have that • But we also know that • So we can also say

  12. Example: Rotation • A dryer rotates at 120 rpm. What distance do your clothes travel during one half hour of drying time in a 70 cm diameter dryer? What angle is swept out?

  13. Example: Rotation • A dryer rotates at 120 rpm. What distance do your clothes travel during one half hour of drying time in a 70 cm diameter dryer? What angle is swept out? • Angle: Dq = w Dt = 120 r/min x 0.5 x 60 min = 120x2p (rad) /min x 60 min/h x 0.5 h = 2.3 x 104 r • Distance: s = Dqr and w = Dq/Dt so s = wDtr • s = 120x2p(rad)/min x 60 min/h x 0.5 h x 0.35 m = 1.3 km

  14. Rotational motion with constant angular acceleration • We will consider cases where a is constant • Definitions of rotational and translational quantities look similar • The kinematic equations describing rotational motion also look similar • Each of the translational kinematic equations has a rotational analogue

  15. Rotational and Translational Kinematic Equations

  16. Constant a motion What is the angular acceleration of a car’s wheels (radius 25 cm) when a car accelerates from 2 m/s to 5 m/s in 8 seconds?

  17. Constant a motion What is the angular acceleration of a car’s wheels (radius 25 cm) when a car accelerates from 2 m/s to 5 m/s in 8 seconds?

  18. A rotating wheel requires 3.00 s to rotate through 37.0 revolutions. Its angular speed at the end of the 3.00-s interval is 98.0 rad/s. What is the constant angular acceleration of the wheel?

  19. A rotating wheel requires 3.00 s to rotate through 37.0 revolutions. Its angular speed at the end of the 3.00-s interval is 98.0 rad/s. What is the constant angular acceleration of the wheel?

  20. Example: Centripetal Acceleration • A 1000 kg car goes around a bend that has a radius of 100 m, travelling at 50 km/h. What is the centripetal force? What keeps the car on the bend? [What keeps the skater in the arc?]

  21. Example: Centripetal Acceleration • A 1000 kg car goes around a bend that has a radius of 100 m, travelling at 50 km/h. What is the centripetal force? What keeps the car on the bend? [What keeps the skater in the arc?] Friction keeps the car and skater on the bend

  22. Car rounding a bend Frictional force of road on tires supplies centripetal force Ifmsbetween road and tires is lowered then frictional force may not be enough to provide centripetal force…car will skid Locking wheels makes things worse as mk < ms Banking of roads at corners reduces the risk of skidding…

  23. Car rounding a bend Horizontal component of the normal force of the road on the car can provide the centripetal force If then no friction is required Ncosq N Nsinq q Fg

  24. An automobile makes a turn whose radius is 150 m. The road is banked at an angle of 18°, and the coefficient of friction between the wheels and road is 0.3. Find the maximum speed for the car to stay on the road without skidding the banked road.

  25. An automobile makes a turn whose radius is 150 m. The road is banked at an angle of 18°, and the coefficient of friction between the wheels and road is 0.3. Find the maximum speed for the car to stay on the road without skidding the banked road. x-component: FNsin q + fs cos q = ma = mv2/R; y-component:FNcos q – fs sin q – mg = 0. FN= mg/(cos q– ms sin q). FNcos q– msFNsin q= mg,

  26. An automobile makes a turn whose radius is 150 m. The road is banked at an angle of 18°, and the coefficient of friction between the wheels and road is 0.3. Find the maximum speed for the car to stay on the road without skidding the banked road. From the x-equation we get:

  27. Rotational Dynamics • Easier to move door atAthan atBusing the same forceF • Moretorqueis exerted atAthan atB hinge B A

  28. Torque • Torque is the rotational analogue of Force • Torque, t, is defined to be WhereFis the force applied tangent to the rotation andris the distance from the axis of rotation t = Fr F r

  29. F r Torque • A general definition of torque is • Units of torque are Nm • Sign convention used with torque • Torque is positive if object tends to rotate CCW • Torque is negative if object tends to rotate CW q t = Fsinq r

  30. Condition for Equilibrium • We know that if an object is in (translational) equilibrium then it does not accelerate. We can say thatSF = 0 • An object in rotational equilibrium does not change its rotational speed. In this case we can say that there is no net torque or in other words that: St = 0

  31. Torque and angular acceleration • An unbalanced torque (t) gives rise to an angular acceleration (a) • We can find an expression analogous to F = ma that relatestand a • We can see that Ft = mat • and Ftr = matr = mr2a (since at = ra) • Therefore Ft r m t = mr2a

  32. Torque and Angular Acceleration t = mr2a Angular acceleration is directly proportional to the net torque, but the constant of proportionality has to do with both the mass of the object and the distance of the object from the axis of rotation – in this case the constant ismr2 This constant is called themoment of inertia. Its symbol isI, and its units arekgm2 Idepends on the arrangement of the rotating system. It might be different when the same mass is rotating about a different axis

  33. Newton’s Second Law for Rotation • We now have that • Where I is a constant related to the distribution of mass in the rotating system • This is a new version of Newton’s second law that applies to rotation t = Ia

  34. A fish takes a line and pulls it with a tension of 15 N for 20 seconds. The spool has a radius of 7.5 cm. If the moment of inertia of the reel is 10 kgm2, through how many rotations does the reel spin? (Assume there is no friction)

  35. A fish takes a line and pulls it with a tension of 15 N for 20 seconds. The spool has a radius of 7.5 cm. If the moment of inertia of the reel is 10 kgm2, through how many rotations does the reel spin? (Assume there is no friction) The torque and corresponding angular acceleration are Using rotational kinematics, the angle through which the spool spins is If we assume the spool starts from rest (ωi= 0) then which is3.5rotations (divide22.5 by2πrotation)

  36. Angular Acceleration andI The angular acceleration reached by a rotating object depends on M, r, (their distribution) and T When objects are rolling under the influence of gravity, only the mass distribution and the radius are important T

  37. Moments of Inertia for Rotating Objects Ifor a small mass mrotating about a point a distance raway is mr2 What is the moment of inertia for an object that is rotating –such as a rolling object? Disc? Sphere? Hoop? Cylinder?

  38. Moments of Inertia for Rotating Objects The total torque on a rotating system is the sum of the torques acting on all particles of the system about the axis of rotation – and sinceais the same for all particles: I Smr2 = m1r12+ m2r22+ m3r32+… Axis of rotation

  39. Continuous Objects To calculate the moment of inertia for continuous objects, we imagine the object to consist of a continuum of very small mass elements dm. Thus the finite sum Σmi r2ibecomes the integral

  40. Moment of Inertia of a Uniform Rod Lets find the moment of inertia of a uniform rod of length L and mass Mabout an axis perpendicular to the rod and through one end. Assume that the rod has negligible thickness. L

  41. Moment of Inertia of a Uniform Rod We choose a mass elementdmat a distancexfrom the axes. The mass per unit length (linear mass density) isλ = M / L

  42. Moment of Inertia of a Uniform Rod dm = λ dx

  43. Example:Moment of Inertia of a Dumbbell A dumbbell consist of point masses 2kg and 1kg attached by a rigid massless rod of length 0.6m. Calculate the rotational inertia of the dumbbell (a) about the axis going through the center of the mass and (b) going through the 2kg mass.

  44. Example:Moment of Inertia of a Dumbbell

  45. Example:Moment of Inertia of a Dumbbell

  46. Moment of Inertia of a Uniform Hoop dm All mass of the hoop Mis at distance r = R from the axis R

  47. Moment of Inertia of a Uniform Disc We expect that Iwill be smaller than MR2since the mass is uniformly distributed from r = 0 to r = R rather than being concentrated atr = R as it is in the hoop. dr r R Each mass element is a hoop of radius rand thicknessdr. Mass per unit area σ = M / A = M /πR2

  48. Moment of Inertia of a Uniform Disc dr r R

  49. Moments of inertia I for Different Mass Arrangements

  50. Moments of inertia Ifor Different Mass Arrangements

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