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Physics I Chapter 10 Dynamics of Rotational Motion.

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Presentation on theme: "Physics I Chapter 10 Dynamics of Rotational Motion."— Presentation transcript:

1 Physics I Chapter 10 Dynamics of Rotational Motion

2 10.1 Torque 1. magnitude of the force 2. the direction of the force 3. the point on the body where the force is applied “line of action” “lever arm (moment arm)”

3 The torque of F with respect to point x. The torque of F about point x. Counterclockwise is defined as positive.. and x. and x SI : Newton-meters but not Joules! T = r x F Example 10.1 TYUS 10.1

4 10.2 Torque and Angular Acceleration for a Rigid Body Only the tangential force component produces a z- component of torque. ΣT z =Iα z for a rigid body Doesn’t apply to a rotating tank of water or a swirling tornado. Only external torques affect the body’s rotation.

5 Weight acts on every particle in the entire body but if g has the same value at all points, we always get the correct torque (about any specified axis) if we assume that all the weight is concentrated at the centre of mass of the body. Weight acts on every particle in the entire body but if g has the same value at all points, we always get the correct torque (about any specified axis) if we assume that all the weight is concentrated at the centre of mass of the body. Example 10.2, 10.3 TYUS 10.2

6 10.3 Rigid-Body Rotation About a Moving Axis = combined translation of the cm and rotation about an axis through the cm K=1/2Mv cm 2 + 1/2I cm ω 2 “Rolling without slipping” v cm =Rω A drag racer first starts to move … Then applying the brakes too heavily … U = Mgy cm Example 10.4, 10.5

7 Combined Translation and Rotation : Dynamics The acceleration of the centre of mass is the same as that of a point mass M acted on by all the external forces on the actual body. ΣF ext = Ma cm ΣT z = I cm α z is valid even when the axis of rotation moves = 1. the axis through the cm must be on the axis of symmetry 2. the axis must not change direction eg. A bicycle Example 10.6, 10.7

8 Rolling Friction perfectly rigid -> no f R piles up in front -> T n not = 0 -> slide : mech.energy will be lost or the rolling body is deformable eg. an automobile tire TYUS 10.3

9 10.4 Work and Power in Rotational Motion dW = T z dθ a constant torque, W = T z Δθ W tot = 1/2Iω 2 2 – 1/2Iω 1 2 P = T z ω z Example 10.8, 10.9 TYUS 10.4

10 10.5 Angular Momentum L = r x p = r x mv a net force F acts on a particle, its velocity and momentum change. dL/dt = r x F = T the perpendicular components on opposite sides of the symmetry axis add up to zero -> L still lies along the axis and L = Iω (for a rigid body)

11 the net force – the internal forces – cancellation - the external forces ΣT = dL/dt for any system of particles, rigid or not…ok! Example 10.10 TYUS 10.5

12 10.6 Conservation of Angular Momentum When the net external torque acting on a system is zero, the total angular momentum of the system is constant (conserved). “pirouetting” The torque of the internal forces can transfer angular momentum from one body to the other, but they can not change the total angular momentum of the system. Example 10.11 – 10.14 TYUS 10.6

13 10.7 Gyroscope and Precession is the rotational analog of uniform circular motion 26,000 years Ω = w r / I ω “Nutation” Example 10.15 TYUS 10.7


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