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Section summary

  • The rotational kinetic energy KE rot size 12{ ital "KE" rSub { size 8{ ital "rot"} } } {} for an object with a moment of inertia I and an angular velocity ω size 12{ω} {} is given by
    KE rot = 1 2 2 . size 12{"KE" rSub { size 8{"rot"} } = { {1} over {2} } Iω rSup { size 8{2} } } {}
  • Helicopters store large amounts of rotational kinetic energy in their blades. This energy must be put into the blades before takeoff and maintained until the end of the flight. The engines do not have enough power to simultaneously provide lift and put significant rotational energy into the blades.
  • Work and energy in rotational motion are completely analogous to work and energy in translational motion.
  • The equation for the work-energy theorem    for rotational motion is,
    net W = 1 2 2 1 2 I ω 0 2 . size 12{"net "W= { {1} over {2} } Iω rSup { size 8{2} } - { {1} over {2} } Iω rSub { size 8{0} rSup { size 8{2} } } } {}

Conceptual questions

The Earth has more rotational kinetic energy now than did the cloud of gas and dust from which it formed. Where did this energy come from?

The figure shows a closed view of a red planet in the sky, with a sun like object seen at the far right and the planet shown here being surrounded by circles of gas and dust.
An immense cloud of rotating gas and dust contracted under the influence of gravity to form the Earth and in the process rotational kinetic energy increased. (credit: NASA)

Problems&Exercises

This problem considers energy and work aspects of [link] —use data from that example as needed. (a) Calculate the rotational kinetic energy in the merry-go-round plus child when they have an angular velocity of 20.0 rpm. (b) Using energy considerations, find the number of revolutions the father will have to push to achieve this angular velocity starting from rest. (c) Again, using energy considerations, calculate the force the father must exert to stop the merry-go-round in two revolutions

(a) 185 J

(b) 0.0785 rev

(c) W = 9 . 81 N size 12{W= {underline {9 "." "81 N"}} } {}

What is the final velocity of a hoop that rolls without slipping down a 5.00-m-high hill, starting from rest?

A baseball pitcher throws the ball in a motion where there is rotation of the forearm about the elbow joint as well as other movements. If the linear velocity of the ball relative to the elbow joint is 20.0 m/s at a distance of 0.480 m from the joint and the moment of inertia of the forearm is 0.500 kg m 2 size 12{0 "." "500"`"kg" cdot m rSup { size 8{2} } } {} , what is the rotational kinetic energy of the forearm?

KE rot = 434 J size 12{ ital "KE" rSub { size 8{ ital "rot"} } = {"434 J"} } {}

A bus contains a 1500 kg flywheel (a disk that has a 0.600 m radius) and has a total mass of 10,000 kg. (a) Calculate the angular velocity the flywheel must have to contain enough energy to take the bus from rest to a speed of 20.0 m/s, assuming 90.0% of the rotational kinetic energy can be transformed into translational energy. (b) How high a hill can the bus climb with this stored energy and still have a speed of 3.00 m/s at the top of the hill? Explicitly show how you follow the steps in the Problem-Solving Strategy for Rotational Energy .

(a) 128 rad/s size 12{ {underline {"128 rad/s"}} } {}

(b) 19.9 m

A ball with an initial velocity of 8.00 m/s rolls up a hill without slipping. Treating the ball as a spherical shell, calculate the vertical height it reaches. (b) Repeat the calculation for the same ball if it slides up the hill without rolling.

Consider two cylinders that start down identical inclines from rest except that one is frictionless. Thus one cylinder rolls without slipping, while the other slides frictionlessly without rolling. They both travel a short distance at the bottom and then start up another incline. (a) Show that they both reach the same height on the other incline, and that this height is equal to their original height. (b) Find the ratio of the time the rolling cylinder takes to reach the height on the second incline to the time the sliding cylinder takes to reach the height on the second incline. (c) Explain why the time for the rolling motion is greater than that for the sliding motion.

What is the moment of inertia of an object that rolls without slipping down a 2.00-m-high incline starting from rest, and has a final velocity of 6.00 m/s? Express the moment of inertia as a multiple of MR 2 size 12{"MR" rSup { size 8{2} } } {} , where M size 12{M} {} is the mass of the object and R size 12{R} {} is its radius.

In softball, the pitcher throws with the arm fully extended (straight at the elbow). In a fast pitch the ball leaves the hand with a speed of 139 km/h. (a) Find the rotational kinetic energy of the pitcher’s arm given its moment of inertia is 0.720 kg m 2 and the ball leaves the hand at a distance of 0.600 m from the pivot at the shoulder. (b) What force did the muscles exert to cause the arm to rotate if their effective perpendicular lever arm is 4.00 cm and the ball is 0.156 kg?

(a) 1.49 kJ

(b) 2.52 × 10 4 N size 12{I= {underline {9 "." "61" times "10" rSup { size 8{3} } " N"}} } {}

Practice Key Terms 2

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Source:  OpenStax, Unit 8 - rotational motion. OpenStax CNX. Feb 22, 2016 Download for free at https://legacy.cnx.org/content/col11970/1.1
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