Parallel-Axis Theorem

Size: px

Start display at page:

Download "Parallel-Axis Theorem"

Gillian Booth
5 years ago
Views:

1 Parae-Axis Theorem In the previous exampes, the axis of rotation coincided with the axis of symmetry of the object For an arbitrary axis, the paraeaxis theorem often simpifies cacuations The theorem states I = I CM + MD 2 I is about any axis parae to the axis through the centre of mass of the object I CM is about the axis through the centre of mass D is the distance from the centre of mass axis to the arbitrary axis

cm + (x cm ) 2 + [(y ) 2 + 2y y cm + (y cm ) 2 ]dm = [(x ) 2 + (y ) 2 ]dm + 2x cm x dm + 2y cm y dm + (x

2 The eement dm is distance r = (x 2 + y 2 ) ½ from the axis I = r 2 dm = (x 2 + y 2 )dm Express the equation above in terms of the CoM coordinates: I = [(x + x cm ) 2 +(y + y cm ) 2 ]dm = [(x ) 2 + 2x x cm + (x cm ) 2 + [(y ) 2 + 2y y cm + (y cm ) 2 ]dm = [(x ) 2 + (y ) 2 ]dm + 2x cm x dm + 2y cm y dm + (x cm 2 + y cm2 ) dm = 0 because the coordinates are reative to the CoM D 2 M Moment of inertia reative to the CoM

3 I = [(x ) 2 + (y ) 2 ]dm + 2x cm x dm + 2y cm y dm + (x cm 2 + y cm2 ) dm Question: Why are these 2 integras zero? Answer: We are integrating x and y and x and y are measured from the CM. Consider one of the integras, e.g. y. The contribution to the quantity inside the integra from distances with y >0 wi be exacty baanced by distances with y <0. In other words, the integra sums to zero. It s the same for x.

4 Moment of Inertia for a Rod Rotating Around One End The moment of inertia of the rod about its centre is ICM = 1 12 D is ½ L ML Therefore, I = I + MD CM L I = ML + M ML 12 = 2 3

Torque Torque, τ, is the tendency of a force to rotate an object about some axis Torque is a vector, but we wi dea with its magnitude first τ = r F sin φ = F d F is the force φ is the ange the force

5 Torque Torque, τ, is the tendency of a force to rotate an object about some axis Torque is a vector, but we wi dea with its magnitude first τ = r F sin φ = F d F is the force φ is the ange the force makes with the horizonta d is the moment arm (or ever arm) of the force The moment arm, d, is the perpendicuar distance from the axis of rotation to a ine drawn aong the direction of the force d = r sin Φ

The horizonta component of the force (F cos φ) has no tendency to produce a rotation Torque wi have direction If the turning tendency of the force is countercockwise, the torque wi be positive If the

6 The horizonta component of the force (F cos φ) has no tendency to produce a rotation Torque wi have direction If the turning tendency of the force is countercockwise, the torque wi be positive If the turning tendency is cockwise, the torque wi be negative Net Torque The force F 1 wi tend to cause a countercockwise rotation about O The force F 2 wi tend to cause a cockwise rotation about O Στ = τ 1 + τ 2 = F 1 d 1 F 2 d 2

7 Torque vs. Force Forces can cause a change in transationa motion Described by Newton s Second Law Forces can cause a change in rotationa motion The effectiveness of this change depends on the force and the moment arm The change in rotationa motion depends on the torque The SI units of torque are N. m Athough torque is a force mutipied by a distance, it is very different from work and energy The units for torque are reported in N. m and not changed to Joues

Torque as a vector We saw before that torque is defined as the product of the appied force and the distance from point of appication to point of rotation. Force and distance are both vectors.

8 Torque as a vector We saw before that torque is defined as the product of the appied force and the distance from point of appication to point of rotation. Force and distance are both vectors. Hence, we need a new way of combining vectors, the cross-product or vector product: A B = C = ABsinθ n ˆ So the magnitude is θ is the ange between A B = ABsinθ A and B The direction of the resutant vector is given by the right hand rue θ

9 To get the direction of the cross-product: 1. Using right hand, point fingers in direction of 2. Cur fingers towards B 3. Thumb now points aong Notes: i) Order of cross-product is important ii) If vectors are parae, cross-product is zero iii) If vectors are at right-anges, cross-product is maximum C C B A B A A B B A C A

10 Torque and Anguar Acceeration Consider a partice of mass m rotating in a circe of radius r under the infuence of tangentia force The tangentia force provides a tangentia acceeration: F t = ma t The radia force, F r causes the partice to move in a circuar path F t

11 The magnitude of the torque produced by F t around the center of the circe is Στ = ΣF t r = (ma t ) r The tangentia acceeration is reated to the anguar acceeration Στ = (ma t )r = (mrα)r = (mr 2 )α Since mr 2 is the moment of inertia of the partice, Στ = Iα The torque is directy proportiona to the anguar acceeration and the constant of proportionaity is the moment of inertia

Consider the object consists of an infinite number of mass eements dm of infinitesima size Each mass eement rotates in a circe about the origin, O Each mass eement has a tangentia acceeration From

12 Consider the object consists of an infinite number of mass eements dm of infinitesima size Each mass eement rotates in a circe about the origin, O Each mass eement has a tangentia acceeration From Newton s Second Law df t = (dm) a t The torque associated with the force and using the anguar acceeration gives dτ = r df t = a t r dm = α r 2 dm Finding the net torque 2 2 τ = αrdm= α rdm This becomes Στ = Ια

13 Faing Smokestack Exampe When a ta smokestack fas over, it often breaks somewhere aong its ength before it hits the ground Each higher portion of the smokestack has a arger tangentia acceeration than the points beow it The shear force due to the tangentia acceeration is greater than the smokestack can withstand The smokestack breaks

14 Torque and Anguar Acceeration, Whee Exampe Cacuate the anguar acceeration of the whee, the inear acceeration of the object and the tension in the cord. Torque on whee is TR, i.e. Στ = Iα =TR α =TR/I Mass moving in straight ine so appy Newton s 2 nd, ΣF= mg - T = ma a = (mg T)/m Three unknowns (α, a & T) but ony two equations need one more Object and whee connected by cord which does not sip, so inear acceeration is same as tangentia a = R α = TR 2 /I = (mg T)/m T = mg/[1 + (mr 2 /I)] a = g/[1 + (I/mR 2 )] α = a/r = g/[r + (I/mR)]

15 Newton s Law of Universa Gravitation Every partice in the Universe attracts every other partice with a force that is directy proportiona to the product of their masses and inversey proportiona to the squared distance between them ( inverse square aw ): F g = G m 1 m 2 r 2 G is the universa gravitationa constant and equas x N m 2 / kg 2 Current We have no theoretica expanation for its vaue Gravity is times weaker than the eectromagnetic force Current known precision on the reative uncertainty is (Rosi et a, Nature 510, (2014) ) In vector form F 12 = G m 1 m 2 r 2 ˆr 12 ΔG G 10 6

16 More About Forces F 12 = F 21 The forces form a Newton s Third Law action-reaction pair Gravitation is a fied force that aways exists between two partices, regardess of the medium between them The force decreases rapidy as distance increases (due to the inverse square aw) F 12 is the force exerted by partice 1 on partice 2 The negative sign in the vector form of the equation indicates that partice 2 is attracted toward partice 1 F 21 is the force exerted by partice 2 on partice 1

17 Gravitationa Force Due to a Distribution of Mass The gravitationa force exerted by a finite-size, sphericay symmetric mass distribution on a partice outside the distribution is the same as if the entire mass of the distribution were concentrated at the center In other words on or above the Earth s surface, a of its mass appears to emanate from a point at its centre (ast part aso true inside spherica distribution) The force exerted by the Earth on a partice of mass m near the surface of the Earth is F g Mm = G R E 2 E

18 negated by matter above here, puing in opposite direction (Birkoff s theorem, in Genera Reativity) If you are here, the gravitation attraction due to matter above is

PHYSICS LOCUS / / d dt. ( vi) mass, m moment of inertia, I. ( ix) linear momentum, p Angular momentum, l p mv l I

PHYSICS LOCUS / / d dt. ( vi) mass, m moment of inertia, I. ( ix) linear momentum, p Angular momentum, l p mv l I 6 n terms of moment of inertia, equation (7.8) can be written as The vector form of the above equation is...(7.9 a)...(7.9 b) The anguar acceeration produced is aong the direction of appied externa torque.