1 Rotational Motion

Circular Motion An object moving in a circle at a constant speed is accelerated Centripetal acceleration depends upon the object’s speed and the radius of the circle Centripetal force causes centripetal acceleration. Skip this assignment:Ch 6 Circular Motion Problems: 12,14-20,49,61-62

FcV a c

Definitions Centripetal Force - a center seeking force that appears to pull an object toward the center of the circle along which it is moving Centripetal Acceleration - acceleration toward the center a c = v 2 /r

Newton’s Law enters in… Since f = ma The centripetal force is determined by substituting a c = v 2 /ra = f/m f c = mv 2 /r

Velocity of an object moving in a circle can be shown as: Revolutions/sec Cycles/sec m/s Degrees/sec AKA: angular velocity

Converting rev/s to m/s 1 rev = 2  r So, if an object completes 3 rev/min in a circle with r = 0.5 m the velocity in m/s is 3 rev 2  (0.5m) 1 min min rev 60 sec V = = 1.57 x m/s

Sample problem

10 Describing Rotary Motion A fraction of a revolution can be measured in degrees, grads or radians A degree is 1/360 of a revolution One revolution is 2  radians One radian is ½  of a revolution

11 Skip this slide Ch8 Assigned problems8/1,6a,11,16,30,34,75,76,82

12 Angular Displacement The Greek letter  is used to represent the angle of revolution The counterclockwise rotation is considered positive while clockwise is negative As the object rotates, the change in angle Is called angular displacement For rotation through an angle, , a point at a distance, r, from the center moves a distance given by d = r 

13 Angular Velocity,  The angular displacement divided by the time required to make the displacement.  =   omega  theta  t Angular velocity can be measured in rad/s, rev/s, degrees/s For Earth, ω E = (2π rad)/(24.0 h)(3600 s/h) = 7.27×10 ─5 rad/s.

14 If an object’s angular velocity is ω, then the linear velocity of a point a distance, r, from the axis of rotation is given by v = rω. The speed at which an object on Earth’s equator moves as a result of Earth’s rotation is given by v = r ω = (6.38×10 6 m) (7.27×10 ─5 rad/s) = 464 m/s.

15 Angular Acceleration,  This is the change in angular velocity divided by the time required to make that change.  =  alpha  t Units: rad/s 2, rev/s 2, degrees/s 2

16 Chart found on p. 199

17 Problem When a machine is switched on, the angular velocity of the motor increases by 10 rad/s for the first 10 seconds before it starts rotating with full speed. What is the angular acceleration of the machine in the first 10 seconds? a. π rad/s 2 b. 1 rad/s 2 c. 100π rad/s 2 d. 100 rad/s 2

18 Answer: B Reason: Angular acceleration is equal to the change in angular velocity divided by the time required to make that change.  =  = 10 rad/s = 1 rad/s 2  t 10 s

19 Rotational Dynamics Torque is a measure of how effectively a force causes rotation. The magnitude of torque is the product of the force and the lever arm. Because force is measured in newtons, and distance is measured in meters, torque is measured in newton-meters (N·m). Torque is represented by the Greek letter tau, .  = Fr sin 

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21 Torque problem A bolt on a car engine needs to be tightened with a torque of 35 N·m. You use a 25-cm-long wrench and pull on the end of the wrench at an angle of 60.0° from the perpendicular. How long is the lever arm, and how much force do you have to exert?

22 Sketch the situation Find the lever arm by extending the force vector backwards until a line that is perpendicular to it intersects the axis of rotation. Label your diagram indicating your known Values Known:Unknown: r = 0.25mL = ?  = 60.0 o F = ?  = 35 Nm

23 Solve for the length of the torque arm L = r sin  = (0.25m) (sin 60.0 o ) = 0.22 m Solve for the force  = F r sin  F =  = 35  m r sin  (0.25m)(sin 60.0 o ) F = 1.6 x 10 2 N

24 Finding Net Force Translational equilibrium: all of the upward forces must equal the downward forces Rotational equilibrium: counterclockwise torque must equal clockwise torque

25 Problem sample…

26 Solution to prior problem 1.75m 6.00kg 4.25 kg F = F so (4.25kg kg)(9.8m/s2) = 100N Let left end = x and right end = 100N - x Choose right end as the axis of rotation (no torque there)

27 Solution continued… 1.75m 58.8N 41.6N cw  = ccw  F 1D1 = F2D2 + F3D3 F1 = F2D2 + F3D3 =41.6(.875)+58.8(.5) D F1 = 37.6N F2 = = 62.4N

28 Moment of Inertia, I - the resistance to rotation For a point mass: I = mr 2 Newton’s 2 nd Law for Rotational Motion:  =  / I

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31 The Center of Mass The center of mass of an object is the point on the object that moves in the same way that a point particle would move. The path of center of mass of the object is a straight line. An object is stable against rollover if its center of mass is above its base.

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34 An object is in equilibrium if there are no net forces exerted on it and if there are no net torques acting on it. Centrifugal “force” and Coriolis “force” are two apparent forces that appear when a rotating object is analyzed from a coordinate system that rotates with it.

Sources All slides are copyrighted and are found in Glencoe Physics Principals and Problems 2005 Edition. 35