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  4. 8.4 - newton's 2nd law applied

0.3 8.4 - newton's 2nd law applied to rotation  (Page 4/7)

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Section summary

  • The farther the force is applied from the pivot, the greater is the angular acceleration; angular acceleration is inversely proportional to mass.
  • If we exert a force F size 12{F} {} on a point mass m size 12{m} {} that is at a distance r size 12{r} {} from a pivot point and because the force is perpendicular to r size 12{r} {} , an acceleration a = F/m size 12{F} {} is obtained in the direction of F size 12{F} {} . We can rearrange this equation such that
    F = ma , size 12{F} {","}

    and then look for ways to relate this expression to expressions for rotational quantities. We note that a = rα size 12{F} {} , and we substitute this expression into F=ma size 12{F} {} , yielding

    F=mrα size 12{F} {}
  • Torque is the turning effectiveness of a force. In this case, because F size 12{F} {} is perpendicular to r size 12{r} {} , torque is simply τ = rF size 12{F} {} . If we multiply both sides of the equation above by r size 12{r} {} , we get torque on the left-hand side. That is,
    rF = mr 2 α size 12{ ital "rF"= ital "mr" rSup { size 8{2} } α} {}

    or

    τ = mr 2 α . size 12{τ= ital "mr" rSup { size 8{2} } α "." } {}
  • The moment of inertia I size 12{I} {} of an object is the sum of MR 2 size 12{ ital "MR" rSup { size 8{2} } } {} for all the point masses of which it is composed. That is,
    I = ∑ mr 2 . size 12{I= sum ital "mr" rSup { size 8{2} } "." } {}
  • The general relationship among torque, moment of inertia, and angular acceleration is
    τ = Iα size 12{τ=Iα} {}

    or

    α = net τ I ⋅ size 12{α= { { ital "net"`τ} over {I} } cdot } {}

Conceptual questions

The moment of inertia of a long rod spun around an axis through one end perpendicular to its length is ML 2 /3 size 12{"ML" rSup { size 8{2} } "/3"} {} . Why is this moment of inertia greater than it would be if you spun a point mass M at the location of the center of mass of the rod (at L / 2 size 12{L/2} {} )? (That would be ML 2 /4 size 12{"ML" rSup { size 8{2} } "/4"} {} .)

Why is the moment of inertia of a hoop that has a mass M and a radius R greater than the moment of inertia of a disk that has the same mass and radius? Why is the moment of inertia of a spherical shell that has a mass M and a radius R greater than that of a solid sphere that has the same mass and radius?

Give an example in which a small force exerts a large torque. Give another example in which a large force exerts a small torque.

While reducing the mass of a racing bike, the greatest benefit is realized from reducing the mass of the tires and wheel rims. Why does this allow a racer to achieve greater accelerations than would an identical reduction in the mass of the bicycle’s frame?

The given figure shows a racing bicycle leaning on a door.
The image shows a side view of a racing bicycle. Can you see evidence in the design of the wheels on this racing bicycle that their moment of inertia has been purposely reduced? (credit: Jesús Rodriguez)

A ball slides up a frictionless ramp. It is then rolled without slipping and with the same initial velocity up another frictionless ramp (with the same slope angle). In which case does it reach a greater height, and why?

Problems&Exercises

Calculate the moment of inertia of a skater given the following information. (a) The 60.0-kg skater is approximated as a cylinder that has a 0.110-m radius. (b) The skater with arms extended is approximately a cylinder that is 52.5 kg, has a 0.110-m radius, and has two 0.900-m-long arms which are 3.75 kg each and extend straight out from the cylinder like rods rotated about their ends.

A soccer player extends her lower leg in a kicking motion by exerting a force with the muscle above the knee in the front of her leg. She produces an angular acceleration of 30.00 rad/ s 2 and her lower leg has a moment of inertia of 0.750 kg ⋅ m 2 size 12{0 "." "750"`"kg" cdot m rSup { size 8{2} } } {} . What is the force exerted by the muscle if its effective perpendicular lever arm is 1.90 cm?

Suppose you exert a force of 180 N tangential to a 0.280-m-radius 75.0-kg grindstone (a solid disk).

(a)What torque is exerted? (b) What is the angular acceleration assuming negligible opposing friction? (c) What is the angular acceleration if there is an opposing frictional force of 20.0 N exerted 1.50 cm from the axis?

(a) 50.4 N ⋅ m

(b) 17.1 rad/s 2 size 12{"17" "." 1``"rad/s" rSup { size 8{2} } } {}

(c) 17.0 rad/s 2 size 12{"17" "." 0``"rad/s" rSup { size 8{2} } } {}

Consider the 12.0 kg motorcycle wheel shown in [link] . Assume it to be approximately an annular ring with an inner radius of 0.280 m and an outer radius of 0.330 m. The motorcycle is on its center stand, so that the wheel can spin freely. (a) If the drive chain exerts a force of 2200 N at a radius of 5.00 cm, what is the angular acceleration of the wheel? (b) What is the tangential acceleration of a point on the outer edge of the tire? (c) How long, starting from rest, does it take to reach an angular velocity of 80.0 rad/s?

The given figure shows the rear wheel of a motorcycle. A force F is indicated by a red arrow pointing leftward at a distance r from its center. Two arrows representing radii R-one and R-two are also indicated. A curved yellow arrow indicates an acceleration alpha and a curved blue arrow indicates an angular velocity omega, both in counter-clockwise direction.
A motorcycle wheel has a moment of inertia approximately that of an annular ring.

Zorch, an archenemy of Superman, decides to slow Earth’s rotation to once per 28.0 h by exerting an opposing force at and parallel to the equator. Superman is not immediately concerned, because he knows Zorch can only exert a force of 4.00 × 10 7 N size 12{4 "." "00" times "10" rSup { size 8{7} } `N} {} (a little greater than a Saturn V rocket’s thrust). How long must Zorch push with this force to accomplish his goal? (This period gives Superman time to devote to other villains.) Explicitly show how you follow the steps found in Problem-Solving Strategy for Rotational Dynamics .

3 . 96 × 10 18 s size 12{3 "." "96" times "10" rSup { size 8{"18"} } `s} {}

or 1.26 × 10 11 y size 12{1 "." "26" times "10" rSup { size 8{"11"} } `y} {}

An automobile engine can produce 200 N ∙ m of torque. Calculate the angular acceleration produced if 95.0% of this torque is applied to the drive shaft, axle, and rear wheels of a car, given the following information. The car is suspended so that the wheels can turn freely. Each wheel acts like a 15.0 kg disk that has a 0.180 m radius. The walls of each tire act like a 2.00-kg annular ring that has inside radius of 0.180 m and outside radius of 0.320 m. The tread of each tire acts like a 10.0-kg hoop of radius 0.330 m. The 14.0-kg axle acts like a rod that has a 2.00-cm radius. The 30.0-kg drive shaft acts like a rod that has a 3.20-cm radius.
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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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