Ch 10 Dynamics of Rotational Motion-final .pdf

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University of Toronto Mississauga
Wagih Ghobriel

cHaPtER 10 Rotational Dynamics According to Newton’s second law, a net force causes an object to have an acceleration.  What causes an object to have an angular acceleration? TORQUE The amount of torque depends on where and in what direction the force is applied, as well as the location of the axis of rotation. Definition of Torque Magnitude of Torque = (Magnitude of the force) x (Lever arm) τ = F  Torque (τ) is defined as the force applied multiplied by the moment arm (or lever arm).  Direction: The torque is positive when the force tends to produce a counterclockwise rotation about the axis, and negative when the force tends to produce a clockwise rotation.  SI Unit of Torque: newton.meter (N·m) Why do acrobats carry long bars? The acrobat controls the net torque acting on her center of gravity by manipulating the pole.Also, the acrobat holds the pole low, moving her center of gravity closer to the rope. Sign Convention A counterclockwise torque is designated as positive(+) A clockwise torque is designated as negative (─) Why internal forces do not contribute to net torque Example: A plumbing problem to solve Example 10.1 • Refer to the worked problem on page 296. Example The Achilles Tendon The tendon exerts a force of magnitude 790 N. Determine the torque (magnitude and direction) of this force about the ankle joint. τ = F   790 N cos55 = 3.6 10 m −2  τ = 720 N 3.6×10 m cos55 =15 N⋅m Center of Gravity (Revisited) Definition of Center of Gravity The center of gravity of a rigid body is the point at which its weight can be considered to act when the torque due to the weight is being calculated. Suppose we have a group of objects, with known weights and centers of gravity, and it is necessary to know the cg fWhen an object has a symmetrical group as a whole: shape and its weight is distributed uniformly, the center of gravity lies at Wx1 1 x + 2 2 ... its geometrical center. xcg W 1 + W 2 + ... Applying an external force by the index finger to provide balance Horizontal board and a weight Horizontal board W x +W x + x = 1 1 2 2 cg W 1W + 2 Example The Center of Gravity of an Arm The horizontal arm is composed of three parts: the upper arm (17 N), the lower arm (11 N), and the hand (4.2 N). Find the center of gravity of the arm relative to the shoulder joint. x = W1 1+W 2 2+ cg W 1+W 2 + 17 N 0.13m + 11N 0.38m + 4.2 N 0.61m ) xcg= = 0.28m 17 N+11N+4.2 N Conceptual Example Overloading a Cargo Plane This accident occurred because the plane was overloaded toward the rear. How did a shift in the center of gravity of the plane cause the accident? Finding the center of gravity of an irregular shape. 10.2 Torque and Angular Acceleration Newton’s Second Law for Rotational Motion About a Fixed Axis F = ma T T τ = FTr aT= rα τ = mr α 2 I = mr is the moment of inertia of the body Notes 2  N2 is valid only in an inertial frame.  Net torque of the internal forces cancel. Although a rigid object possesses a unique total mass, it does not have a unique moment of inertia. 2 2 τ1= m 1 1α ∑ τ = ∑ (mr α 2 τ2= m 2 2α Net external  torque Moment of inertia τN= m N Nα Rotational analog of newton’s second law for a rigid body rotating about a fixed axis Moment of  Angular  Net externaltorque=inertia  acceleration     Requirement:Angular acceleration ∑ τ = I α must be expressed in radians/s . I= (mr 2 ∑ Example: Rotating masses With respect to the variables given in the figure, the equation for the angular acceleration α is: (a) rF/2 mR (b) RF/2 mr (c) rF/3 mR (d) RF/3 mr (e)2 /3 mr Reminder Example The Moment of Inertial Depends on Where the Axis Is. Two particles each have mass m and are fixed at the ends of a thin rigid rod. The length of the rod is L. Find the moment of inertia when this object rotates relative to an axis that is perpendicular to the rod at (a) one end and (b) the center. (a) I = ∑ mr = m r 1 1r = 2 2 +m L )2 ( )2 2 I = mL m 1 m =2m r = 0 r = L 1 2 (b) I = ∑ (mr = m r +m r = m L 2 +m L 22 ( )2 1 2 1 1 2 2 I = m2 r = L 2 r = L 2 1 2 Reminder Reminder Example: Hoisting a Crate The combined moment of inertia of the dual pulley is 50.0 kg·m 2. The crate weighs 4420 N. Atension of 2150 N is maintained in the cable attached to the motor. Find the angular acceleration of the dual pulley. equal F = T −mg = ma ∑ τ =T1 1  2 2α ∑ y 2 y ay= 2α T 2 mg +ma y T1 1− mg m+ y)2 = Iα ay = 2α T − mg +m α  = Iα 1 1 2 2 T −mg α = 1 1 2 2 I +m 2 (2150 N 0.600m − 451kg 9.80m s 2)(0.200m ) 2 = 2 2 = 6.3rad s 46.0kg m + 451kg 0.200m ) The yo-yo rotates on a moving axis. A bowling ball rotates on a moving axis. Refer to worked example on pages 302 and 303. 10.3 Rotational Work, Energy and Power in Rotational Motion W = Fs = Fr θ s = rθ τ = Fr W =τθ Definition of Rotational Work The rotational work done by a constant torque in turning an object through an angleΔθ is W R = τ. Requirement:  SI Unit of Rotational Work:joule (J)s. Work can be done by a constant torque. Example 10.6 follows the generic situation to the right. 2 2 2 KE = 2v =Tmr 2 v = rω T 1 2 2 1 2 2 1 2 KE = ∑ (2mr ω = ) (2 ∑ mr ω = Iω 2 Definition of Rotational Kinetic Energy The rotational kinetic energy of a rigid rotating object is 2 KE =RIω2 Requirement:  The angular speed must be expressed in rad/s.  SI Unit of Rotational Kinetic Energy:joule (J) Power in Rotationalal Motion ∆∆ θ P = = τ = τω ∆∆t 10.4 Angular Momentum Definition of Angular Momentum The angular momentum L of a body rotating about a fixed axis is the product of the body’s moment of inertia and its angular velocity with respect to that axis: L = Iω Requirement:  The angular speed must be expressed in rad/s.  SI Unit of Angular Momentum:kg·m Newton’s second law of motion Conservation Impulse-momentum Sum of average of angular  Impulse of a torque. t
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