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Gogsey
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1. A wedge of mass M slides on a frictionless horizontal surface; a block of mass m slides on the frictionless
inclined surface of the block, which is inclined at angle theta to the horizontal.
a) Using variables X for the position of the wedge, and s for the location of the
block on the wedge, write out the lagrangian (remember to find speeds relative to
an inertial frame).
b) Write out the lagrange equations of motion. Solve to find the time for the block to reach the bottom, if it
begins a distance s0 up the wedge, and the system is released from rest.
Just want to confirm the answer to part b).
So for an inclined plane we have t=sqrt(2So/x1dotdot) where x1 is the acceleration of the for the block taking into account the effect of the wedge on the motion of the block, which was found to be:
x1dotdot = gsin(theta)/(1 - (mcos^2(theta))/(m+M))
So I get t = sqrt(2So - (2So mcos^2(theta))/(M+M) / gsin(theta)).
Can this be simplified, or is this fine, assuming its correct? Just seems really ugly.
2. The wedge in the previous question is glued to the floor of an accelerating truck, so that its position is given by
X = ct^2 . The block is free to slide on the surface of the wedge.
a) How many degrees of freedom are there?
b) Use the lagrangian method again to find the motion s(t) of the block.
c) For what value of the constant c does the block remain at constant s?
Just for finding the lagrangian. Is it better to leave the speed on the truck as x2dot, and then sub in the 2ct at the end? Its just that if you sub it in at the beginning when trying to obtain an expression for the kinetic energy of the wedge/truck and the block, you wont' have any x2dot in the expression, where x2dot is the velocity of the wedge due to the trucks acceleration in the horizontal direction and since x2 is ct^2, x2dot is 2ct.
inclined surface of the block, which is inclined at angle theta to the horizontal.
a) Using variables X for the position of the wedge, and s for the location of the
block on the wedge, write out the lagrangian (remember to find speeds relative to
an inertial frame).
b) Write out the lagrange equations of motion. Solve to find the time for the block to reach the bottom, if it
begins a distance s0 up the wedge, and the system is released from rest.
Just want to confirm the answer to part b).
So for an inclined plane we have t=sqrt(2So/x1dotdot) where x1 is the acceleration of the for the block taking into account the effect of the wedge on the motion of the block, which was found to be:
x1dotdot = gsin(theta)/(1 - (mcos^2(theta))/(m+M))
So I get t = sqrt(2So - (2So mcos^2(theta))/(M+M) / gsin(theta)).
Can this be simplified, or is this fine, assuming its correct? Just seems really ugly.
2. The wedge in the previous question is glued to the floor of an accelerating truck, so that its position is given by
X = ct^2 . The block is free to slide on the surface of the wedge.
a) How many degrees of freedom are there?
b) Use the lagrangian method again to find the motion s(t) of the block.
c) For what value of the constant c does the block remain at constant s?
Just for finding the lagrangian. Is it better to leave the speed on the truck as x2dot, and then sub in the 2ct at the end? Its just that if you sub it in at the beginning when trying to obtain an expression for the kinetic energy of the wedge/truck and the block, you wont' have any x2dot in the expression, where x2dot is the velocity of the wedge due to the trucks acceleration in the horizontal direction and since x2 is ct^2, x2dot is 2ct.
