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Thursday, October 29, 2015

± The Work of a Force
Learning Goal:
To be able to calculate the work done by a force.
If an external force, F, acts on a rigid body, the work, UF, done by the force when it moves along a path s is defined as
UF=rFdr=sFcosθds
where θ is the angle between the force and the object's direction of motion and r is the position vector that points to the object as it moves along path s.
 
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Part A
Which of the following forces do work?
  • A compressed spring that is allowed to expand.
Part B
A crate of mass m = 54.0 kg slides a distance l = 4.60 m down a ramp at an angle θ = 25.0 . At the bottom of the ramp is a spring that has a spring constant k = 680 N/m . (Figure 1) What is s, the distance the spring will compress when the crate comes to rest? The ramp is smooth enough that the friction is negligible.
 
 
 
 
Part C
When a satellite is launched, its orbital radius is usually determined far in advance because the amount of energy required to change its orbit is large. How much work, W, would a rocket thruster on a satellite of mass m = 610 kg need to do to change the satellite's orbital radius from r1 = 2.84×107 m to r2 = 4.16×107 m ? (Figure 2) The Earth's radius is RE = 6.37×106 m . Assume that both orbits are circular.
 
 
 
 
 
 
± The Work of a Couple
Learning Goal:
To be able to calculate the work done by a couple on a rigid body.
When a body subjected to a couple moment, M, undergoes general planar motion, the two couple forces do work only when the body undergoes a rotation. When the body rotates in the plane through a finite angle θ (measured in radians) from θ1 to θ2, the work of a couple is
UM=θ2θ1Mdθ
If the couple moment, M, has a constant magnitude, then the work is reduced to UM=M(θ2θ1)



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Part A
Each blade on a wind turbine is engineered to produce an effective force through its center of mass that is perpendicular to its long axis. (Figure 1) The magnitude of the force produced by each blade in a wind of speed v is F=(40.0 kg/s)v. If the center of mass of each blade is r = 12.4 m , what is U1, the total work done by the wind turbine's blades, in a wind of speed v = 8.50 m/s after 10 revolutions?

Part B
A brake system is tested by rotating a tire and measuring the number of rotations required for the brake system to bring the tire to a stop. (Figure 2) The tire's radius is R = 50.0 cm and the brake system's radius is r = 17.9 cm . A moment of M = 16.9 Nm is applied to the tire for 5 rotations before the brake system is applied. The brake system is composed of two pads that are pushed out against the drum with a force that increases as the tire rotates and is described by F=(10.0θ) N. If the coefficient of kinetic friction between the brake pads and the outer ring of the brake system is μk = 0.550, how many rotations, n, will the tire go through before coming to a stop?




Reading Question 18.04
Part A
Static friction at A does work, and static friction at B does no work
Pulling a String Adds Energy to a Wheel
Learning Goal:
To apply the work-energy theorem to solve a planar kinetic problem.
(Figure 1)
A bicycle wheel, with moment of inertia I and radius r, is mounted on a fixed, frictionless axle, with a light string wound around its rim. The wheel is rotating counterclockwise with angular velocity ω1, when at time t=0 someone starts pulling the string with a force of magnitude F. Assume that the string does not slip on the wheel.
 
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Part A
Suppose that after a certain time, tL, the string has been pulled through a distance L. What is ω2, the final rotational speed of the wheel?
 
Part B
What is P, the instantaneous power delivered to the wheel via the force F at time t=0?