Unformatted text preview: Chapter 11 Rota,on and angular momentum Torque We already know, torque on a body is deﬁned as the vector product of the posi,on vector of the point at which the force acts and the force: τ =r×F Both magnitude and direc,on of the torque is given by this equa,on. Thus the magnitude is: τ = r F sin θ = rF sin θ = r⊥ F = rF⊥ The direc,on is given by the right hand rule. Both magnitude and direc,on can be calculated using the method of calcula,ng cross product of two vectors: ˆ ˆ ˆk i j ˆ ˆ ˆj ˆ τ = r × F = ( xi + yˆ + zk ) × ( F i + F ˆ + F k ) = x y z j
x y z Fx Fy Fz Angular momentum In transla,onal mo,on, we know that a net force produces change in linear momentum of a par,cle. What is the rota,onal equivalent? First, consider a par,cle of linear momentum p = mv and acted upon by a net force F. Let r be the posi,on vector of the par,cle. Net torque on the par,cle about the origin is: τ =r×F dp Using Newton’s second law: τ = r × dt θ dr × p (which is zero since dr/dt = v and p = mv are parallel, and does not Add the term dt make any diﬀerence): dp dr τ =r× + ×p dt dt Angular momentum (contd…) The previous discussion leads to d τ = ( r × p ) ....(1) dt
We see that torque is equal to rate of change of ( r × p ) . This sound like Newton’s second law (net force is equal to rate of change of momentum). The vector quan,ty ( r × p ) is called angular momentum of the par,cle (about the axis of rota,on) and is denoted by L. Its SI unit is kgm2s. L=r×p The magnitude is L = mvrsinθ. The direc,on is given by right hand rule. Or we can calculate both magnitude and direc,on using the rule of vector product. Note that when linear momentum is parallel to the posi,on vector, angular momentum is zero With this deﬁni,on, equa,on(1) becomes: dL τ= dt That is, the torque ac/ng on a par/cle is equal to the rate of change of the par/cle’s angular momentum. Angular momentum of a system of par,cles and rigid object For a system of par,cles: Then, τ net dL = dt L = ∑ Li
i For a rigid body (assume r and v are in xy plane): L = ∫ dL = ˆ ˆ ( r × dp ) = ∫ ( r × v )dm = k ∫ r (ω r ) dm = kω ∫ r 2 dm ∫ The magnitude of angular momentum is therefore given by L = Iω
Compare with p = mv. Conserva,on of angular momentum We already know: dL If net torque ac,ng on a system is zero, then = 0 ⇒ L = const . dt τ net dL = dt That is, The the total angular momentum of a system is constant in both magnitude and direc>on if the net external torque ac>ng on the system is zero, that is, if the system is isolated. This statement is called the principle of conserva4on of angular momentum. For an isolated rigid body rota,ng about an axis conserva,on of angular momentum means: L f = Li ⇒ I f ω f = I iω i Example: A rod with M = 1 kg and length l = 1 m is pivoted about its center and is ini>ally oriented in the x
direc>on. A 50 g clump of clay hits the right end of the rod at a velocity of 2 m/s making an angle of 300 with the x
axis as shown. The clay s>cks to the rod. What is the resul>ng angular velocity of the rod ? 300 Angular momentum just before the impact: Li = Li , rod + Li , clay = 0 + mr × v
l Li = m v sin 150° = (0.05 )(0.5 )(2 ) sin 150° = 0.025 kg m 2 /s 2 v
300 1500 r Angular momentum aZer the impact: ⎛1 2⎞ L f = ( I1 + I 2 )ω = ⎜ Ml 2 + m ( l / 2 ) ⎟ ω = (0.096 kgm 2 )ω ⎝ 12 ⎠ Angular momentum conserva,on: L f = Li ⇒ ω = 0.26 rads −1 Also see Example 11.9 in the textbook. Example: Problem 29 (Ch 11 textbook) Chalkboard stuﬀ Precession of gyroscopes Let’s see a small you tube clip: gyroscope (h^p://www.youtube.com/watch?v=cquvA_IpEsA&feature=related) A gyroscope is a^ached to a shaZ on which it can pivot. If it does not spin, it will fall by rota,ng down about the shaZ. If it is spinning, it will not fall; but instead it will rotate about the shaZ, perpendicularly to the direc,on of the gravita,onal force. Why ? Image of a coated gyroscope rotor and matching housings. (h^p:// www.nasa.gov/mission_pages/gpb/ gpb_003.html) Precession of gyroscopes (contd…) Forces ac,ng on the gyroscope: its weight mg (gravity) and the normal force at the pivot. Only gravity gives torque. r r τ mg Pivot point τ = r × F = r × mg The direc,on of the torque is as shown in the ﬁgure above. If the gyroscope is not spinning, this simply means that it will fall down rota,ng about the shaZ as shown. If the gyroscope is spinning, the torque causes angular momentum to change ( dL = τ dt Ini,ally, the gyroscope has an angular momentum of This means dL ⊥ L , which in turn means that the change in angular momentum is not due to the change in magnitude of L, but to change in its direc,on. This make the gyroscope precess its axis of rota,on, in the horizontal plane in this example. ): Precession of gyroscopes (contd…) What is the angular speed of precession? We know: Spinning top seen from above: dL L dϕ Since dL is perpendicular to L it will cause a change in angle of L equal to: dϕ = dL Mgr dt = L Iω Dividing by dt we get the precession speed (about the the rota,on axis of the gyroscope) : dϕ ωP =
dt → ωP = Mgr Iω Or, ωP = τ Iω Example: Earth precesses with a period of 26,000 years due to >dal force from the moon & the sun. What is the approximate magnitude of the torque causing this precession ? Since ω P =
τ ⇒ τ = ω P Iω Iω If this torque is applied to the surface of earth, the corresponding force is 8.5 x 1015N! ...
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 Force, Momentum, Trigraph, the00, τ Iω

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