Kinematics
A machine is a device that accepts energy in some available form and utilizes it to do a type of work. Energy, work, or power has to be transferred from one mechanical part to another to run a machine. While the transfer of energy between two machine parts, those two parts experience a relative motion with each other. Studying such relative motions is termed kinematics.
Kinetic Energy and Work-Energy Theorem
In physics, work is the product of the net force in direction of the displacement and the magnitude of this displacement or it can also be defined as the energy transfer of an object when it is moved for a distance due to the forces acting on it in the direction of displacement and perpendicular to the displacement which is called the normal force. Energy is the capacity of any object doing work. The SI unit of work is joule and energy is Joule. This principle follows the second law of Newton's law of motion where the net force causes the acceleration of an object. The force of gravity which is downward force and the normal force acting on an object which is perpendicular to the object are equal in magnitude but opposite to the direction, so while determining the net force, these two components cancel out. The net force is the horizontal component of the force and in our explanation, we consider everything as frictionless surface since friction should also be calculated while called the work-energy component of the object. The two most basics of energy classification are potential energy and kinetic energy. There are various kinds of kinetic energy like chemical, mechanical, thermal, nuclear, electrical, radiant energy, and so on. The work is done when there is a change in energy and it mainly depends on the application of force and movement of the object. Let us say how much work is needed to lift a 5kg ball 5m high. Work is mathematically represented as Force ×Displacement. So it will be 5kg times the gravitational constant on earth and the distance moved by the object. Wnet=Fnet times Displacement.
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An electric motor exerts a constant torque of τ=10N⋅mτ=10N⋅m on a grindstone mounted on its shaft; the moment of inertia of the grindstone is I=2.0kg⋅m2I=2.0kg⋅m2. If the system starts from rest, find the work done by the motor in 8.0 ss and the kinetic energy at the end of this time. What was the average power delivered by the motor? |
SOLUTIONSET UP AND SOLVE We want to use W=τΔθW=τΔθ to find the total work done and then divide that by the time interval Δt=8.0sΔt=8.0s to find the average power (work per unit time). But first we need to find the total angle ΔθΔθ (in radians) through which the grindstone turns. The The angular acceleration is α=τI=10N⋅m2.0kg⋅m2=5.0rad/s2α=τI=10N⋅m2.0kg⋅m2=5.0rad/s2 For constant angular acceleration, θ=θ0+ω0t+12αt2=0+0+12(5.0rad/s2)(8s)2=160radθ=θ0+ω0t+12αt2=0+0+12(5.0rad/s2)(8s)2=160rad The total work is W=τΔθ=(10N⋅m)(160rad)=1600JW=τΔθ=(10N⋅m)(160rad)=1600J Finally, the average power is the total work divided by the time interval: Pav=WΔt=1600J8.0s=200J/s=200WPav=WΔt=1600J8.0s=200J/s=200W REFLECT We can check the energy relationships. The ω=αt=(5.0rad/s2)(8.0s)=40rad/sω=αt=(5.0rad/s2)(8.0s)=40rad/s The grindstone's kinetic energy at this time is K=12Iω2=12(2.0kg⋅m2)(40rad/s)2=1600JK=12Iω2=12(2.0kg⋅m2)(40rad/s)2=1600J This amount equals the total work done, as of course it must. Note that the power is not constant during this process. The torque is constant, but as the angular velocity increases, so does the rate of doing work. The instantaneous power, τωτω, increases linearly with time—from zero, when ω=0ω=0, to 400 WW, with an average value of 200 WW. Suppose we now turn off the motor and apply a brake to bring the grindstone from its greatest angular speed of 50 rad/srad/s to a stop. If the brake generates a constant torque of 30 N⋅mN⋅m , find the total work done (including signs) needed to stop the grindstone. Express your answer in joules.
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Find the average power (including signs) needed to stop the grindstone in Part A
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