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Example

Example. A 40-kg mass placed 1.25 m on the opposite side of the support point balances a mass of 25 kg, placed (x) m from the support point of a uniform beam. What is the value of the unknown distance x?.

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Example

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  1. Example A 40-kg mass placed 1.25 m on the opposite side of the support point balances a mass of 25 kg, placed (x) m from the support point of a uniform beam. What is the value of the unknown distance x?

  2. A large seed initially at rest explodes into two pieces which move off. Which of these could be possible paths the two pieces would take? (II) (III) (I)

  3. A ball is projected straight up. Which graph shows the total energy of the ball as a function of time? (D) (C) (B) (A) t t t t (E) (F) t t

  4. Chapter 8 Torque ()and Angular Momentum (L)

  5. Rotational Inertia (I) Recall: From Newton’s 2nd law of linear motion: Fnet = ma = m(v-vo)/t mass (m) = measure of inertia of an object. = measure of how difficult it is to change linear velocity (v) of an object. The larger the mass, the more difficult it is to change its velocity v. For rotation, moment of inertial or rotational inertia (I) = measure of how difficult it is to change angular velocity () of an object.

  6. Rotational Inertia (I) • Rotational inertia (I) = measure of how difficult it is to change angular velocity () of an object. • Rotational inertia (I) – Also called moment of inertia: • Depends on mass, m. (I  m). • Depends on square of radius of rotation (I  r2) • Depends on how mass m is distributed. Same mass. Different radius. Which one easier to get to start rotating?

  7. Rotational Inertia For a point object, I = mr2 Point object = one whose size is small c.f. radius r m r Units of I = kg-m2

  8. Rotational Inertia For a point object, I = mr2 For many discrete point objects with different shapes rotated about the same axis, Total rotational inertia = sum of I for each object. Itotal = miri2 m1 r1

  9. Rotational Inertia • For a point object, I = mr2 • For objects of comparable size to radius r, the moment of in inertia depends on distribution of mass.

  10. Rotational Inertia Table

  11. Rotational Inertia (I) Mass of earth = 5.975 x 1024 kg Mean radius = 6.37 x 106 m Earth-sun distance = 1.5 1011 m What is the rotational inertia of the earth’s spin about its axis?

  12. Example m2 = 5 kg m1 = 3 kg Two solid spherical balls are joined together by a 4-meter long steel rod. If they are spun about a vertical axis passing through their center of mass, find the total rotational inertia. [Hint, treat the spinning balls as point objects with I = mR2].

  13. Rotational Inertia (I)

  14. Rotational Kinematics

  15. Comment on axes and sign(i.e. what is positive and negative) Whenever we talk about rotation, it is implied that there is a rotation “axis”. This is usually called the “z” axis (we usually omit the z subscript for simplicity). Counter-clockwise (increasing q) is usuallycalled positive. Clockwise (decreasing q) is usuallycalled negative. +w z

  16. Example You and a friend are playing on a merry-go-round. You stand at the outer edge of the merry-go-round and your friend stands halfway between the outer edge and the center. Assume the rotation rate of the merry-go-round is constant. Who has the greatest angular velocity? 1. You do 2. Your friend does 3. Same

  17. Example Who has the greatest tangential velocity (v)? 1. You do 2. Your friend does 3. Same

  18. v m r Rotational Kinetic Energy Mass m in rotational motion. Its rotational inertia, I = mr2 Since it is moving, it has kinetic energy. K = ½ mv2 From v = r, K = ½ mr22 = ½ (mr2) 2 = ½I2

  19. Kinetic Energy of Rotating Disk Consider a disk with radius R and mass M, spinning with angular frequency w • Each “piece” of disk has speed vi=wri • Each “piece” has kinetic energy • Ki = ½ mi v2 • = ½ miw2 ri2 • Combine all the pieces • SKi = S ½ miw2 ri2 • = ½ (S mi ri2) w2 • = ½ I w2 mi ri

  20. Rotational Kinetic Energy (Krot) A rigid object spinning about a fixed axis (pure rotational motion) has rotational kinetic energy Krot = ½I2 [eg, spinning wheel] If the rigid object is moving (sliding) with velocity v without any spin (ie pure translational motion), it has only translational kinetic energy K = ½ mv2 [eg, skidding wheel]

  21. Rotational Kinetic Energy (Krot) If the rigid object is spinning with angular velocity  while its center of mass moves linearly with velocity vcm, it has both translational and rotational kinetic energy. Its total kinetic energy will be Ktot = Ktran + Krot ½ mvcm2 + ½ I2 [eg, wheel rolling normally on the ground, ball rolling on the ground]

  22. Ktran = ½ m v2 Linear Motion • Krot = ½ I w2 Rotational Motion

  23. Example Who has the greater kinetic energy? (Assume masses are equal) 1. You do 2. Your friend does 3. Same

  24. Example A 10-kg hollow cylindrical shell rolls on the ground at a linear velocity of 5 m/s. Find its total kinetic energy. 50 cm 50 cm

  25. Torque (t) • Force (F) is responsible for change of linear velocity. Net force results in linear acceleration. • Torque (t) is responsible for change in angular velocity. Net torque results in angular acceleration.

  26. Torque r  r F t = rF = (rsin )F = rFsin r =rsin = lever arm (moment arm) Torque = lever arm x force Unit: m-N

  27. r  F t = rF = rsinF Larger lever arm r, larger torque t if F stays unchanged. 2. Larger force F, larger torque t if lever arm stays unchanged. 3. If r = 0, t = 0. 4. If  = 0, t = 0. 5. If  = 90o, t = rF

  28. Torque Torque is a vector quantity. Counter clockwise rotation, torque is positive F F Clockwise rotation, torque is negative

  29. A string is tied to a doorknob 0.80 m from the hinge as shown in the figure. At the instant shown, the force applied to the string is 5.0 N. What is the torque on the door? hinge

  30. Equilibrium • Recall, translational equilibrium – Fnet = 0. • Fnet = 0, object at rest – static equilibrium. • Fnet = 0, object moving with constant velocity, - dynamic equilibrium. • Rotational equilibrium, net = 0. • An object in equilibrium means it is in both translational equilibrium and rotational equilibrium. • Conditions for equilibrium: Fnet = 0 and net = 0

  31. Equilibrium Conditions for equilibrium: Fnet = 0 and net = 0 • Torque can be calculated about any desired axis. • A judicious choice of axis helps. • Choose axis at a point through which an unknown force acts so that its torque does not appear in the equation.

  32. Center of Gravity • Is the point through which the force of gravity acts. • Essentially the same as center of mass.

  33. M2 = 60 kg M1 = 50 kg 1.5 m 2.5 m pivot • The figure above is a snapshot of two masses on a light beam placed on a pivot. • What is the net torque about the pivot? • To be in rotational equilibrium, what should have been the value of M2?

  34. Example The picture below shows two people lifting a heavy log. Which of the two people is supporting the greatest weight? • The person on the left is supporting the greatest weight • The person on the right is supporting the greatest weight • They are supporting the same weight

  35. L R Example The picture below shows two people (L & R) lifting a 20-kg log. The length of the log is 4.0 m and L is 1.0 m from the left end while R is holding from the right end. Find how much force each of them have to exert.

  36. Newton’s 2nd Law • Newton’s 2nd law for linear motion: Fnet = ma Eg. What force is required to change velocity of a 6 kg solid spherical object of radius 30 cm object from 15 m/s to 25 m/s in 5 seconds? • Newton’s 2nd law for rotation: net = I Eg. What torque is required to change angular velocity of a 6 kg solid spherical object of radius 30 cm object from 15 rad/s to 25 rad/s in 5 seconds? [Assume spin about the axis]

  37. Work Done by Torque • Recall, work done by constant force W = F x cos q Also, Wnet = K = ½ mvf2 – ½ mvi2 • For rotation, work done by constant torque: W = Ftangential s = Ftangential r q = t q Also, Wnet = Krot = ½ If2 – ½ Ii2 Power, P = W/t = t q/t = tw

  38. Work Done by Torque • A 182-kg flywheel has an effective radius of 0.62 m. • How much torque will take it from rest to a rotational speed of 120 rpm within 30.0 s? • How much work will this take?

  39. Rolling Object • A rolling object posses both Ktran (½ mv2CM) and Krot (½ I2) E1 = U1 + K1 = mgh + 0 h E2 = U2 + K2 0 + ½ mv2 + ½ I2 • If it came down rolling: mgh = ½ mvA2 + ½ I2 • If it came down only skidding, mgh = ½ mvB2

  40. Angular Momentum (L) Recall: Linear Momentum: p = mv Angular Momentum, L = I Angular momentum of a rigid object rotating about a fixed axis. L is a vector quantity. Units = kg-m2/s Direction: If rotation is CCW, L is upward. If rotation is CW, L is downward.

  41. Conservation of Angular Momentum If net external torque acting on a rotating object is zero, total angular momentum will be conserved. If net = 0, Lbefore = Lafter OR Iii = Iff EG: Skater spinning with her arms extended away from her body. Lbefore = Iii = (mri2)i She then decides to pull her arms close to her body so that her spinning radius becomes smaller (rf < ri) and Lafter = Iff = (mrf2)f

  42. Conservation of Angular Momentum Since no external torque acts on her, principle of conservation of angular momentum will hold: Lbefore = Lafter or Iii = Iff or (mri2)i = (mrf2)f Since ri > rf, i < f ie, the skater will spin faster by just folding her arms.

  43. Example • A 1000-kg merry-go-round of radius 8.0 m is spinning at a uniform speed of 25 rpm. If a 150-kg mass is gently placed at the edge, what will be the final spin speed of the merry-go-round?

  44. Example 4 3 2 Axis 1 • Which of the forces in the figure below produces the largest torque about the rotation axis indicated? Assume all the four forces have equal magnitudes. • 2 • 3 • 4 • The torques are equal

  45. A 250-kg merry-go-round of radius 2.2 m is spinning at a uniform angular speed of 3.5 rad/s. Calculate the magnitude of its angular momentum. Use I = ½ MR2 • 605 kg-m2/s • 1.93 x 103 kg-m2/s • 7.41 x 103 kg-m2/s • 2.12 x 103 kg-m2/s • 963 kg-m2/s

  46. A 30.0-cm long wrench is used to generate a torque of 14.6 N-m at a bolt. A force of 70.0 N is applied at the end of the wrench at an angle of  to the wrench. The angle  is • 44o • 0.4o • 9.2o • 3.6o

  47. A 3.2-kg solid sphere of diameter 28 cm spins at 5.0 rad/s about an axis passing through its center. How much work is needed to bring it to rest within 15 sec? [Use I = 2/5 x mr2] • 0.025 J • 1.25 J • 1.25x104 J • 3.14 x 103 J • 0.31 J

  48. Abel and Brian support a 68.0-kg uniform bar that is 4.0 m long. Abel holds the bar 1.0 m from one end, while Brian holds from the other end. How much force does Abel have to exert? • 444 N • 45.3 N • 1.33x103 N • 333 N • 222 N

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