Tiêu đề 5. Rotational Motion.pdf ✅

Rotational Motion 87 5 Rotational Motion QUICK LOOK Rotational Kinematic equations are 0 ω ω= + at α = angular acceleration 1 2 0 2 θ ω= +t at ω0 = initial angular speed 2 2 0 ω ω αθ = + 2 ω = angularspeed after time t θ = angular displacement Figure: 5.1 When a figure skater draws her arms and a leg inward, she reduces the distance between the axis of rotation and some of her mass, reducing her moment of inertia. Since angular momentum is conserved, her rotational velocity must increase to compensate. Angular momentum conservation plays a very important role in all figure skating routines. In uniform horizontal circular motion, speed, kinetic energy, angular velocity, angular acceleration and angular momentum are constant. For a uniform circular path the direction of linear velocity is tangential, the direction of angular velocity is along the axis of circular path. The direction of acceleration and force is towards centre. Figure: 5.2 Angular velocity 2 2 n t t θ π ω π ∆ = = = ∆ . If r is radius of circle, linear velocity v r = ω Centripetal acceleration 2 v 2 a r r = = ω Centripetal force 2 mv 2 F mr r = = ω Variables of Circular Motion Displacement: The change of position vector of the displacement ∆r of the particle form position A to the position B is given by referring the figure. Figure: 5.3 1 2 ∆ = − r r r 2 1 ∆ = ∆ = − r r r r 2 2 1 1 1 2 ∆ = + − r r r r r 2 cosθ Putting 2 2 r r r = = we obtain 2 2 ∆ = + − ⋅ r r r r r 2 cosθ ( ) 2 2 2 2 1 cos 2 2sin 2 r r r θ θ   ∆ = − =     ⇒ 2 sin 2 r r θ ∆ = Distance: The distance covered by the particle during the time t is given as d = length of the arc AB = rθ Ratio of distance and displacement cosec 2 2 2 sin 2 d r r r θ θ θ θ   = =   ∆   Figure: 5.4 Change in velocity: 2 1 ∆ = − v v v or 2 1 ∆ = − v v v ⇒ 2 2 1 2 1 2 ∆ = + − v v v v v 2 cosθ For uniform circular motion 1 2 v v v = = B O A 2 v 1 v θ B O A 2 v 1 v θ 1 r 2 r ω π = 2 f v r ω = S r θ = θ f = frequency in revolutions v v r Axis of rotation at ac at affects magnitude ac affects directions r v ω′ L = Iω L = I′ω′ ω
88 Quick Revision NCERT-PHYSICS So ( ) 2 2 1 cos 2sin 2 v v θ ∆ = − = θ The direction of ∆v is shown in figure that can be given as 180 90 2 2 θ θ φ ° −   = = ° −     If two particles are moving on same circle or different coplanar concentric circle in same direction with different uniform angular speeds ωA and ωB respectively, the angular velocity of B relative to A will be ω ω ω rel B A = − So the time taken by one to complete on revolution around O with respect to the other (i.e., time in which B completes one more or less revolution around O than A) 1 2 2 1 1 2 2 2 rel T T T T T π π ω ω ω = = = − − 2 as T π ω   =     Special Case: If , 0 ω ω ω B A rel = = and so T = ∞, particles will maintain their position relative to each other. This is what actually happens in case of geostationary satellite (ω ω 1 2 = = constant) Circular Motion: d d dr r dt dt dt υ ω ω → → → = × + × Here, , dv a dt = (Resultant acceleration) a r = × + × α ω υ α ω = → dt d (Angular acceleration) t c a a a = + Figure: 5.5 Table 5.1: Tangential and Centripetal Acceleration Centripetal acceleration Tangential acceleration Net acceleration Type f motion ac = 0 at = 0 ac = 0 Uniform translatory motion ac = 0 at ≠ 0 a = at Accelerated translatory motion ac ≠ 0 at = 0 a = ac Uniform circular motion ac ≠ 0 at ≠ 0 2 2 c t a a a = + Non-uniform circular motion Figure: 5.6 Vertical Circular Motion Critical velocity at uppermost point v rg c = ( ) Critical speed of lowest point ( ) 5 c v rg = Tension at any point 2 cos mv T mg r = + θ Figure: 5.7 Figure: 5.8 Velocity at any point on vertical loop 2 v u gh = − 2 2 = − − u gh 2 (1 cos ) θ h l l l = − = − cos 1 cos θ θ ( ) Where l is the length of the string. Tension at any point on vertical loop 2 cos mv T mg l − = θ 2 cos mv T mg l = + θ ( ) 2 2 3cos m T u gl l = − −   θ   [As ( ) 2 v u gl = − − 2 1 cosθ ] O m P θ l θ T 2 cos mv mg l θ + C D B A v O P θ l C D B A u θ = 90° θ = 270° θ = 0° θ = 180° mg mg mg 2 mv T mg r = − 2 mv T mg r = + 2 mv T r = 2 mv T r = mg Motion of the CM plus... cm v cm −v cm v v = 0 2 cm v these two velocity vectors sum to give zero velocity at bottom The bottom of wheel is at rest! But only for an instant ... motion about the CM equals... ... motion of individual point on the wheel P θ a a a O
Rotational Motion 89 Table 5.2: Velocity and Tension in a Vertical Loop Position Angle Velocity Tension A 0° u 2 mu mg l + B 90° 2 u gl − 2 2 2 mu mg l − C 180° 2 u gl − 4 2 5 mu mg l − D 270° 2 u gl − 2 2 2 mu mg l − It is clear from the table that: T T T A B C > > and T T B D = 3 T T mg A B − = 6 T T mg A C − = and 3 T T mg B C − = Table 5.3: Various Conditions for Vertical Motion Velocity at lowest point Condition u gl A > 5 Tension in the string will not be zero at any of the point and body will continue the circular motion. u gl A = 5 Tension at highest point C will be zero and body will just complete the circle. 2 5 gl u gl < < A Particle will not follow circular motion. Tension in string become zero somewhere between points B and C whereas velocity remain positive. Particle leaves circular path and follow parabolic trajectory 2 A u gl = Both velocity and tension in the string becomes zero at B and particle will oscillate along semicircular path. 2 A u gl < Velocity of particle becomes zero between A and B but tension will not be zero and the particle will oscillate about the point A Table 5.4: Different Variables in Vertical Loop Quantity Point A Point B Point C Point D Point P Linear velocity (v) 5gl 3gl gl 3gl gl( ) 3 2cos + θ Angular velocity (ω) 5g l 3g l g l 3g l ( ) 3 2cos g l + θ Tension in String (T) 6mg 3mg 0 3mg 3 1 cos m g ( + θ ) Kinetic Energy (KE) 5 2 mgl 3 2 mgl 1 2 mgl 3 2 mgl 5 2 mgl mgl − (1 cos ) − θ Potential energy (PE) 0 mgl 2mgl mgl m gl (1 cos − θ ) Total Energy (TE) 5 2 mgl 5 2 mgl 5 2 mgl 5 2 mgl 5 2 mgl Banking of Curve Force equation at maximum speed v, at threshold of sliding up incline. 2 sin cos x v F m N N r Σ = = + θ μ θ 0 sin cos Σ = = ⋅ − F N N mg y θ μ θ Solving this pair of equations for the maximum speed v gives: ( ) max sin cos cos sin rg V θ μ θ θ μ θ + = − The limiting cases are: max Frictionless case V rg = tanθ ; max Flat roadway V rg = μ Figure: 5.9 Rail track The banking angle θ of curved tack is 2 tan v rg θ = Maximum velocity for no skidding is max v rg = μ Maximum speed for no overturning max 2 rdg v h = where h is height of centre of gravity; d = separation between inner and outer wheels. When a vehicle tends to overturn, the inner wheels leave the ground first. Figure: 5.10 Torque τ = ×r F or τ θ = rF sin Equation of motion of rotating rigid body τ = I α Moment of inertia about given axis 2 I mr = Σ . If k is radius of gyration I = Mk2 , M = Σm = total mass of body Theorem of parallel axes 2 G I I Mr = + Figure: 5.11 IG I r Normal force N Centripetal force θ Rail θ mg θ θ N mg 2 net v F m r = f N = μ r = radius of curvature of curve. y x
90 Quick Revision NCERT-PHYSICS Theorem of perpendicular axes z x y I I I = + 2 2 2 , , x y z ∆ = ∆ ∆ = ∆ ∆ = ∆ I my I mx I mr Then ( ) 2 2 x y ∆ + ∆ = ∆ + I I m x y Since 2 2 2 r x y = + it follows that 2 x y ∆ + ∆ = ∆ I I mr z = ∆I Figure: 5.12 Since this is true for any mass element then x y z I I I + = Kinetic energy of rotation 1 2 2 = Iω Kinetic energy of rolling body 1 1 2 2 2 2 cm = + I mv ω Figure: 5.13 Work energy theorem in rotational motion 2 2 2 1 1 1 2 2 W I I = − ω ω Rolling Motion on Inclined Plane Figure: 5.14 Velocity 2 2 2 sin 1 gs v k R θ = + , Acceleration 2 2 sin 1 g a k R θ = + S = distance on inclined plane, k = radius of gyration R = radius of symmetrical body and θ the inclination angle. Frictional force on a body rolling on inclined plane 2 2 sin 1 Mg k R θ = + Rotationally Symmetric Bodies Rolling Down a Slope Condition of no slipping v r = ω Conservation of energy 1 1 2 2 sin 2 2 mv I m g x E + − = ω α Acceleration 2 sin ( / ) dv mg dt m I r α = + Table 5.5: Comparison of Rolling, Sliding and Falling Motions Physical Quantity Rolling motion Sliding motion Falling motion Velocity 2 F gh v β = 2 2 2 sin 1 ( / ) gs k R θ = + v gh s = 2 = 2 2 sin gs θ β θ θ = = ° 1, , 2 F v gh = Acceleration sin r a g θ β   =     2 2 sin 1 ( / ) g k R θ = + sin s a g = θ F a g = Time of descent 1 2 sin R h t g β θ   =     2 2 2 1 1 sin k h R g β θ     +   = 1 2 sin S h t θ g = 2 sin h g θ = 2 F h t g = Table 5.6: Acceleration, Velocity and Time of Descend for Different Bodies Rolling Down an Inclined Plane Body 2 sin 1 g a I Mr θ =     +   2 2 1 gh V I Mr =   +     1 sin t θ = 2 2 1 gh I V g Mr   = +     Solid sphere 5 sin 7 g θ 10 7 gh 1 14 sin 5 h θ g Hollow sphere 3 sin 5 g θ 6 5 gh 1 10 sin 3 h θ g Disc 2 sin 3 g θ 4 3 g h 1 3 sin h θ g Cylinder 2 sin 3 g θ 4 3 g h 1 3 sin h θ g Hollow cylinder 1 sin 2 g θ g h 1 4 sin h θ g Ring 1 sin 2 g θ g h 1 4 sin h θ g Table 5.7: Ratios of Rotational KE (KR), K = translational (KT) and total KE(K) of Different Bodies Body 2 ( ) I MK = 2 K 2 2 2 2 1 2 1 2 R T k Mv K r K Mv ⋅ = 2 2 k r = 2 2 2 2 1 2 1 1 2 T Mv K K k Mv r =     +   2 2 2 1 ( / ) k k r = + 2 2 2 2 1 R k K r K k r = + Ring/Hollo w cylinder 2 r 1 1 1 2 1 2 Hollow sphere 2 2 3 r 2 3 3 5 2 5 Disc/solid cylinder 2 2 r 1 2 2 3 1 3 Solid sphere 2 2 5 r 2 5 5 7 2 7 θ h S v I = is related to the moment of inertia about the point of contact by the parallel axis theorem ω = is related to V cm if it is rolling without slipping ω r X Z Y