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Year 13 Physics Rotation & Circular Motion Rotation When either a rigid body or a particle rotates about some fixed point, we can describe the motion in a similar fashion to an object moving in a straight line. v r θ Consider a particle moving in a circle, Radius r, speed v. In time t, the radius Sweeps out an angle θ. Some important definitions. Angular Displacement (angle) An angle θ (measured in radians) is defined as the ration of length of arc d, to the radius, r. d r d r Units: radians (rad) d Angular Velocity We define angular velocity, ω (omega) as: change in angular displacement time t Units: rad.s-1 Angular Acceleration If an particle moving in a circle speeds up, its angular velocity also increases. We define angular acceleration, α (alpha) as: change in angular velocity time t Units: rad.s-2 Relationship between angular and linear quantities As d = rθ we can write: v d r r. r t t t Similarly, as: a i.e. v r r. r t t t v r a r Equations of Rotational Motion Consider the particle shown moving in a circle radius r. At time = 0 it has angular velocity ωi. It undergoes constant angular acceleration α to reach ωf in time t. In this time the radius line rotates through an angle θ. As for the linear situation it is possible to derive the following equations. The only difference being that: ω replaces v α replaces a θ replaces i ωf r θ α d ωi Rotational f i t (i f )t 2 Linear v f vi at d (vi v f )t 2 i t 12 t 2 d vi t 12 at 2 2f i2 2 v 2f vi2 2ad Angular Momentum An object moving in a straight line has linear momentum given by: p = mv. We define angular momentum of a moving particle about some fixed point P as: L mvr m r Note: r is the distance measured from P to the ‘line of action’ of the velocity at right angles as shown. For a particle moving in a circle this would always be the radius of the circle. P v Torque A torque is a ‘turning’ force. In order to cause an object to turn or rotate, the must be some pivot point P, about which it turns and the force must act some distance away from that point. F d P NB: The distance d is measured from the pivot at right angles to the ‘line of action’ of the force F. We define: Torque = Force x distance. Fd Units are: Newton.metres (N.m) Rotational Dynamics For the particle, mass m moving in a circle, it has linear energy given by: E 12 mv 2 but v = rω so this becomes: E 12 mr 2 2 The particle has angular momentum. L = mvr Using the same substitution, this becomes: L mr 2 v r θ m s If a force acts in the direction of motion, the particle will accelerate. F=ma Using the substitutions: Fr a r This becomes: mr 2 Rotational Inertia Each of the last three equations contain the term mr2. We define this quantity for a PARTICLE as Rotational Inertia I. I mr 2 NB. For more complex shapes we must use: I r 2 dm These three equations become: ER 12 I 2 L I I NB. These are the rotational analogies to their linear counterparts: I replaces m ω replaces v α replaces a τ replaces F L replaces p ER replaces Ek This is called Rotational kinetic energy) Example: A particle, mass 2.0kg moves in a circle, radius 5.0m, with constant speed. It completes 1revolution every 4.0s. Find: 2r 2 5 7.9m.s 1 T 4 (a) Its speed. v (b) Its angular velocity. (c) Its rotational inertia. The object is a particle so: v 7.9 1.6rad .s 1 r 5 I mr 2 2 52 50kg.m2 (d) Its rotational energy. ER 12 I 2 0.5 50 1.6 2 62 J Rotational Inertia for ‘Real’ Objects The value of I for rotating bodies other than a particle is usually found by considering energy considerations or using the conservation of angular momentum. Solid cylinder, mass M, radius R I 12 MR 2 Solid sphere, mass M, radius R I 52 MR 2 Conservation of Angular Momentum When the nett torque on a system is zero, angular momentum is always conserved. I11 I 22 This is best shown with an example. An ice skater goes into a spin with arms extended. In this position, she has a rotational inertia I = 6.8kg.m2. In this position she turns 8 revolutions in 5s. Find her angular velocity and angular momentum in this position. 8 2 10rad .s 1 t 5 L I 6.8 10 68kg.m 2 .s 1 Next, she pulls her arms close to her body. What effect will this have on her rotational inertia, and why? What will happen to her angular velocity, and why? It will reduce the value of I as the mass is closer to the axis of rotation. As angular momentum, Iω is conserved, if I decreases, angular velocity ω will increase. In this new position, it is found that she completes12 revolutions in the next 5s. Find her new angular velocity, hence her new rotational inertia. 12 2 15rad .s 1 t 5 I 22 68kg.m 2 .s 1 I2 68 4.5kg.m 2 15 Find her energy before and energy after she tucks her arms in. Ei 12 I112 0.5 6.8 10 2 340 J E f 12 I 222 0.5 4.5 152 506J How has this increase in energy come about? In order to bring her arms in towards the axis of rotation, she must do work. The work done is equal to the change in energy. Circular Motion When a particle moves in a circle with constant speed it accelerates. This is because its velocity (a vector) is continually changing, as its direction is always changing. v m r ar This centripetal (or radial) acceleration ar is directed towards the centre of the circle. v2 ar r Centripetal (radial) Force This is the force required to hold an object in a circle. Fr = mar. mv2 Fr r Frequency and Period The time taken for a rotating object to complete one revolution is the PERIOD, T. The number of revolutions in 1 second is the FREQUENCY, f. These two quantities are related : 1 T f Which leads to: f 1 T In one revolution: v circumference period 1revolution period v 2r 2rf T 2 2f T Example: The Conical Pendulum θ Consider a mass on a string suspended as shown moving in a horizontal circle. l The forces acting on the mass are: Tension v r m W These forces can be shown on a vector diagram. FR is the resultant force acting towards the centre. W Tension FR Find an expression for the period of rotation. Note that: r l sin FR mg tan v 2r T tan Substitute into Hence: sin cos mv 2 FR r T 2 l cos g