Physics 325 HW 9 Solutions

# Physics 325 HW 9 Solutions - Physics 325 Homework 9...

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Physics 325 Homework 9 – Solutions – Fall 2010 1) A hoop of mass m and radius R rolls without slipping down an inclined plane of mass M , that makes an angle θ with the horizontal. Find the Lagrange equations and the integrals of the motion if the plane can slide without friction along a horizontal surface. [ Hint : Recall that the kinetic energy of rotation can be written as T rot = 1 2 I ! 2 , with I the moment of inertia. For a hoop, I = mR 2 .] Solution : We set up a coordinate system as is shown in the figure: We place the origin of the x - y coordinate system fixed on the surface that holds the inclined plane. The position of the incline will be measured as x 1 , y 1 ( ) and the position of the hoop will be measured as x 2 , y 2 ( ) . We will measure the position of the hoop along the plane as s measured from the bottom of the plane. Note we have defined the height and base of the incline as h and d respectively for convenience, they will not show up in the result. We begin by noting that five variables are required to specify the motion of the hoop, four spatial dimensions ( x 1 , y 1 , x 2 , y 2 ) and one angular dimension ( φ ) to specify the orientation of the hoop. We have three constraints: The inclined plane remain on the table: y 2 =0. The hoop moves on the surface of the incline plane. This relates the x2 and y2 coordinates or, alternatively, relates x 2 and y 2 to s . This can be written as: sin = y 2 / s . (0.1) The rotational motion of the hoop is related to the translational motion. This constraint can be written as: s = R " ! s = R ! . (0.2) Five coordinates with three constraints gives two degrees of freedom. We choose coordinates x 1 and s to specify our system. The potential energy function can be written as: U y ( ) = mgy 2 = mgs sin (0.3) R g hoop, mass m I = mR 2 h d incline mass M y x s

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where we have defined the potential function so that U (0)=0. We have additionally written the potential energy in terms of s based upon equation (0.1). The kinetic energy of the inclined plane is: T plane = 1 2 M ! x 1 2 (0.4) The kinetic energy of the hoop is going to have a contributions from the translational motion of the hoop along the inclined plane, the translational motion of the inclined plane along the surface, and a contribution from rotational motion: T hoop = 1 2 m ! v i ! v + 1 2 I " ! 2 (0.5) We can write the velocity of the hoop as the sum of the motion along the inclined plane plus the motion of the inclined plane: ! v = " s ˆ s + " x 1 ˆ i = ! " s cos " + " x 1 ( ) ˆ i + " s sin ˆ j (0.6) Note: as the hoop rolls down the incline ( x 2 increasing) s is decreasing. We now write the kinetic energy of the hoop as: T hoop = 1 2 m ! ! s cos + ! x 1 ( ) 2 + ! s 2 sin 2 # \$ % ( + 1 2 I ! ) 2
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## This note was uploaded on 10/06/2011 for the course PHYS 325 taught by Professor Staff during the Fall '08 term at University of Illinois, Urbana Champaign.

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Physics 325 HW 9 Solutions - Physics 325 Homework 9...

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