PS113

Chapter 9

Rotational Dynamics

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The effects of forces and torques on the motion of rigid objects • The mass of most rigid objects is spread out and not concentrated at a single point. There are three kinds of motion a particle can have as it moves through space: 1. Pure translation (no rotation) 2. Curvilinear motion 3. Pure rotation (no translation) • Today, we’re going to introduce the new dynamical quantity called torque. Before we do this, we need to introduce two new concepts: 1. The line of action is an extended line drawn colinear with the force. 2. The lever arm is the distance ` of closes approach between the line of action and the axis of the rotation, measured on a line that is perpendicular to both. • The torque is represented by the symbol τ (Greek letter tau) whose magnitude is: Torque = (Magnitude of the force) × (Lever arm)

τ = F` • Example: Pushing on a door 1

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Problem 3:

You are installing a new spark plug in your car, and the manual specifies that it be tightened to a torque that has a magnitude of 45 N·m. Using the data in the drawing, determine the magnitude F of the force that you must exert on the wrench. ` = 0.28 m (sin 50o ) Answer: 210 N

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Rigid objects in equilibrium • There are two kinds of equilibrium in physics–translational equilibP P rium ( Fx = 0, Fy = 0) as described by Newton’s first law, and rotational equilibrium: X τ =0 (2) • Equilibrium of a rigid body: A rigid body is in equilibrium if it has zero translational acceleration and zero angular acceleration. In equilibrium, the sum of the externally applied forces is zero, and the sum of the externally applied torques is zero.

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Center of Gravity • It is often important to know the torque produced by the weight of an extended body. In the previous example, it was necessary to determine the torques caused by the weight of the ladder and the man on the ladder. In this case, the weight is considered to act at a definite point for the purpose of calculating the torque. This point is called the center of gravity • Definition of the center of gravity: The center of gravity of a rigid body is the point at which its weight can be considered to act when calculating the torque due to the weight. W1 x 1 + W2 x 2 + · · · xcg

=

=

(W1 + W2 + · · ·) xcg

W1 x1 + W2 x2 + · · · W1 + W2 + · · ·

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• Example: A 1-kg meter stick has a 200 g mass placed at 30 cm and a 250 g mass placed 90 cm. If the c.g. of the meter stick is located at 50 cm and the mass of the meter stick is 50 g, where is the center of gravity of the combined system located? Problem 12: The drawing shows a person whose weight is W = 584 N doing push-ups. Find the normal force exerted by the floor on each hand and each foot, assuming that the person holds this position. feet → c.g. = 0.840 m, and c.g. → hands = 0.410 m

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Newton’s second law for rotational motion about a fixed axis • We will now explore how torque is used in Newton’s 2nd law of motion. More specifically, we will look at the torque applied to a rigid body rotating about a fixed axis. Recall that every point in a rigid body shares the same angular velocity (ω = v/r) and angular acceleration (α = aT /r) • The torque applied to one point on a rigid body is: τ1 = r1 F1 sin θ = r1 FT where F1 sin θ is the force (FT ) perpendicular to r1 , thus making r1 the lever arm. Continuing on, we find that: τ1 = r1 FT = r1 (maT ) = r1 m (r1 α) = (m1 r12 ) α

Since a rigid body is made up of many point-like particles, we can imagine that each particle experience a local torque to some external force (e.g., gravity), and we can write: τ1 + τ2 + τ3 + · · · = (m1 r12 )α + (m2 r22 )α + (m3 r32 )α + · · · τ1 + τ2 + τ3 + · · · = (m1 r12 + m2 r22 + m3 r32 + · · ·)α 6

Finally, we can write that: τ1 + τ2 + τ3 + · · · =

X

τi =

N X

(mi ri2 )α = Iα

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i=1

or X

Problem 34:

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τext = Iα

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A ceiling fan is turned on and a net torque of 1.8 N·m is applied to the blades. The blades have a total moment of inertia of 0.22 kg·m2 . What is the angular acceleration of the blades?

Rotational work and energy • We saw before that the work performed to cause translational motion was W = (F cos θ) · s. In many cases, the force was in the direction of motion, so this equation reduced to W = F · s. In rotational dynamics this equation becomes: W =τ ·θ

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where τ is the external torque applied to the system, and θ is the angular displacement. • We can now express the Work-Energy Theorem for rotational motion as: Wrot = τ · θ = ∆KErot where ∆KErot is the rotational kinetic energy created by the torque. • The total work done by forces that cause translation as well as rotation on a rigid body is: Wtotal = Wtrans + Wrot = ∆KEtrans + ∆KErot 1 1 1 1 = mv 2 − mvo2 + Iω 2 − Iω 2 2 2 2 2 !

Wtotal

!

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Demonstration: Suppose a hoop and disk, both of mass M and radius R, are released at rest from the top of an incline. Which one will arrive at the bottom of the incline first? v u q u4 vdisk = t gh vhoop = gh 3 What makes the difference in velocity? It’s how the mass is distributed? Notice that the velocities do not depend on the mass or the radius–but only on how the mass is distributed.

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Problem 48:

Calculate the kinetic energy that the earth has because of (a) its rotation about its own axis and (b) its motion around the sun. Assume that the earth is a uniform sphere and that its path around the sun is circular. For comparison, the total energy used in the United States in one year is about 9.3 × 1019 J.

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Angular Momentum • Just as linear momentum is conserved when the Fext = 0, so is P angular momentum conserved when the τext = 0. P

• We define the angular momentum L of a body rotating about a fixed axis as the product of the body’s moment of inertia I and its angular velocity ω with respect to that axis: L = Iω

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where ω is expressed in [rad/s] and the SI Unit of angular momentum is kg · m2 /s • For a satellite in an elliptical orbit about the earth, the point of closest approach is called the perigee while the furthest point is called the apogee. Their respective distances from the earth are labeled (rP ) and (rA ). • Requiring conservation of angular momentum at the perigee and apogee, we can write the following: 

LP = LA

⇒

IP ωP = IA ωA

⇒



v mr2  P  P

rP





=

v mr2  A  A

rA

Finally, we cancel some terms in the above equation and write the relationship between the perihelion (rP ) and the aphelion (rA ) as: rP vP = rA vA Notice that the mass of the satellite is of no consequence. Also notice that the earth and the satellite are considered to be a closed system where there are no external torques acting on the system, therefore, we can presume that angular momentum is conserved.

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Problem 59:

Two disks are rotating about the same axis. Disk A has a moment of inertia of 3.4 kg·m2 and an angular velocity of +7.2 rad/s. Disk B is rotating with an angular velocity of -9.8 rad/s. The two disks are then linked together without the aid of any external torques, so that they rotate as a single unit with an angular velocity of -2.4 rad/s. The axis of rotation for this unit is the same as that for the separate disks. What is the moment of inertia of disk B.

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