Chapter 8 Rotational Motion Rotational Motion Rigid body

Chapter 8 Rotational Motion

Rotational Motion Rigid body = a body with a definite shape that doesn’t change, so that the particles composing it stays in fixed positions relative to one another.

Angular Quantities In purely rotational motion, all points on the object move in circles around the axis of rotation (“O”). The radius of the circle is r. All points on a straight line drawn through the axis move through the same angle in the same time. The angle θ in radians is defined: where l is the arc length. (some times given as s also)

We see that when the disk rotates so that the arc length (s) equals the length of the radius of the disk (r), angle (θ) will equal 1 radian. The resulting equation is s = rθ θ = s/r If the disk rotates through one complete revolution, then s equals the entire circumference and θ equals 2π radians s = rθ 2πr = rθ θ = 2π radians where the unit of a radian represents the dimensionless measure of the ratio of the circle's arc length to its radius Since one complete revolution equals 360º, we now have the conversion that 360º = 2π radians 1 radian = 180/π or approximately 57. 3º

Angular Quantities Angular displacement: The average angular velocity is defined as the total angular displacement divided by time:

Angular Quantities The angular acceleration is the rate at which the angular velocity changes with time:

Angular Quantities Every point on a rotating body has an angular velocity ω and a linear velocity v. They are related:

Angular Quantities Therefore, objects farther from the axis of rotation will move faster. Buzz Lightyear! Ladybugs

Angular Quantities If the angular velocity of a rotating object changes, it has a tangential acceleration: Even if the angular velocity is constant, each point on the object has a centripetal acceleration:

Angular Quantities Here is the correspondence between linear and rotational quantities:

Constant Angular Acceleration The equations of motion for constant angular acceleration are the same as those for linear motion, with the substitution of the angular quantities for the linear ones.

Problem 1 An old phonograph record revolves at 45 rpm. What is its angular Once the motor is turned off, velocity in rad/sec? it takes 0. 75 seconds to come ω = 45 rev/min = 4. 71 rad/sec to a stop. What is its average angular acceleration? givens: ωf = 0, ωo = 4. 71 rad/sec, t = 0. 75 seconds. using the equation ωf = ωo + αt we can determine that α = (0 - 4. 71)/0. 75 = -6. 28 rad/sec 2

How many revolutions did it make while coming to a stop? Combining α = ω/t and θ = ωot +½αt 2 we get: θ = ½(ωf + ωo)t we can determine that θ = ½ (0 + 4. 71) 0. 75 θ = 1. 77 radians since there are 2π radians in every revolution, θ = 0. 281 rev.

A fan that is turning at 10 rev/min speeds up to 25 rev/min in 10 seconds. b. If the tip of one blade is 30 cm from the center, what is the final tangential velocity of the tip? a. How many revolutions does the blade require to alter using the equation its speed? v = rω allows us to determine that using the equation v = (0. 30)(2. 63) θ = ½(ωf + ωo)t v = 0. 789 m/sec we can determine that θ = ½(1. 05 + 2. 63)10 θ = 18. 4 radians since there are 2π radians in every revolution, θ = 2. 92 rev

Consider a standard analog wall clock with a second hand, minute hand, and hour hand a. Calculate the angular velocity of the second hand of a clock ω = 1 rev/min = 0. 105 rad/sec b. If the second hand is 8" long (there are 2. 54 cm in every inch), what is the linear velocity of the tip of the second hand? using the equation v = rω allows us to determine that v = (0. 203)(0. 105) v = 0. 0213 m/sec

Two wheels are connected by a common cord. One wheel has a radius of 30 cm, the other has a radius of 10 cm When the small wheel is revolving at 10 rev/min, how fast is the larger wheel rotating? Since the two wheels share the same tangential velocity, their angular velocities will be inversely proportional to their radii. vlarge=vsmall rlargeωlarge= rsmallωsmall ωlarge = (rsmall/rlarge) ωsmall ωlarge = (0. 10/0. 30)(10) = 3. 33 rev/min

A rotor turning at 1200 rev/min has a diameter of 5 cm. As it turns, a string is to be wound onto its rim How long of a piece of string will be wrapped in 10 seconds? since the wheel is turning at a constant angular velocity, we can use the equations θ = ωt s = rθ substituting gives us the equation s = r(ωt) and we can calculate the amount of string wrapped around the exterior of the rotor s = (0. 025)(1200)(0. 105)(10) s = 31. 5 meters

Rolling Motion (Without Slipping) In (a), a wheel is rolling without slipping. The point P, touching the ground, is instantaneously at rest, and the center moves with velocity v. In (b) the same wheel is seen from a reference frame where C is at rest. Now point P is moving with velocity –v. The linear speed of the wheel is related to its angular speed:

Example A Bicycle slows down from 8. 4 m/s to rest over a distance of 115 m. a) Find the angular velocity of the wheels when the bike tis traveling 8. 4 m/s. b) The total number of revolutions each wheel makes before coming to a stop. c) The angular acceleration of the wheel? d) The time it took to stop?

Torque To make an object start rotating, a force is needed; the position and direction of the force matter as well. The perpendicular distance from the axis of rotation to the line along which the force acts is called the lever arm.

Units for Torque is proportional to the magnitude of F and to the distance r from the axis. Thus, a tentative formula might be: t = Fr Units: N m or lb ft t = (40 N)(0. 60 m) = 24. 0 N m, cw t = 24. 0 N m, cw 6 cm 40 N

Torque A longer lever arm is very helpful in rotating objects.

Torque Here, the lever arm for FA is the distance from the knob to the hinge; the lever arm for FD is zero; and the lever arm for FC is as shown.

8 -4 Torque The torque is defined as:

Example 1: An 80 -N force acts at the end of a 12 -cm wrench as shown. Find the torque. • Extend line of action, draw, calculate r. r = 12 cm sin 600 = 10. 4 cm t = (80 N)(0. 104 m) = 8. 31 N m

Alternate: An 80 -N force acts at the end of a 12 -cm wrench as shown. Find the torque. positive 12 cm Resolve 80 -N force into components as shown. Note from figure: rx = 0 and ry = 12 cm t = (69. 3 N)(0. 12 m) t = 8. 31 N m as before

Example 2: Find resultant torque about axis A for the arrangement shown below: Find t due to each force. Consider 20 -N force first: r = (4 m) sin = 2. 00 m negative 30 N r 300 6 m 40 N 300 t = Fr = (20 N)(2 m) = 40 N m, cw 2 m 20 N 300 A 4 m The torque about A is clockwise and negative. t 20 = -40 N m

Example 2 (Cont. ): Next we find torque due to 30 -N force about same axis A. Find t due to each force. Consider 30 -N force next. r = (8 m) sin = 4. 00 m 300 t = Fr = (30 N)(4 m) = 120 N m, cw r negative 30 N 300 20 N 300 6 m 40 N 2 m A 4 m The torque about A is clockwise and negative. t 30 = -120 N m

Example 2 (Cont. ): Finally, we consider the torque due to the 40 -N force. Find t due to each force. Consider 40 -N force next: r = (2 m) sin 900 = 2. 00 m t = Fr = (40 N)(2 m) = 80 N m, ccw positive 30 N r 300 2 m 6 m 40 N 20 N 300 A 4 m The torque about A is CCW and positive. t 40 = +80 N m

Example 2 (Conclusion): Find resultant torque about axis A for the arrangement shown below: Resultant torque is the sum of individual torques. 20 N 300 300 2 m 6 m 40 N A 4 m t. R = t 20 + t 20 = -40 N m -120 N m + 80 N m t. R = - 80 N m Clockwise

Rank the wrenches from the one that is experiencing the most torque to the one experiencing the least

Where should the 100 g mass be placed to balance the system?

Challenge Problem: What is the mass of the bolt in the picture below

Challenge Problem: Where do you need to put the support to balance the meters tick shown below? Ignore the mass of the meter-stick.

Rotational Dynamics; Torque and Rotational Inertia Knowing that , we see that This is for a single point mass; what about an extended object? As the angular acceleration is the same for the whole object, we can write:

Rotational Dynamics; Torque and Rotational Inertia The quantity is called the rotational inertia of an object. The distribution of mass matters here – these two objects have the same mass, but the on the left has a greater rotational inertia, as so much of its mass is far from the axis of rotation.

Rotational Dynamics; Torque and Rotational Inertia The rotational inertia of an object depends not only on its mass distribution but also the location of the axis of rotation – compare (f) and (g), for example.

A merry go round accelerates from rest to 0. 68 rad/s in 34 s. Assuming the merry go round is a uniform disk of radius 7. 0 m and mass 31, 000 kg, calculate the net torque required to accelerate it. I = ½ MR 2.

Rotational Kinetic Energy The kinetic energy of a rotating object is given by By substituting the rotational quantities, we find that the rotational kinetic energy can be written: A object that has both translational and rotational motion also has both translational and rotational kinetic energy:

Rotational Kinetic Energy When using conservation of energy, both rotational and translational kinetic energy must be taken into account. All these objects have the same potential energy at the top, but the time it takes them to get down the incline depends on how much rotational inertia they have.

A bowling ball of mass 7. 25 kg and radius 10. 8 cm rolls without slipping down a lane at 3. 1 m/s. Calculate the total Kinetic Energy of the ball. The lane is 18 m long. Hope many rotations does the ball make on the way down the lane?

Rotational Kinetic Energy The torque does work as it moves the wheel through an angle θ:

Angular Momentum and Its Conservation In analogy with linear momentum, we can define angular momentum L: We can then write the total torque as being the rate of change of angular momentum. If the net torque on an object is zero, the total angular momentum is constant.

Angular Momentum and Its Conservation Therefore, systems that can change their rotational inertia through internal forces will also change their rate of rotation: video

What is the angular momentum of a 2. 8 kg uniform cylindrical grinding wheel of radius 28 cm rotating at 1300 rpm? How much torque is required to stop it in 6. 0 s?