A homogeneous cylinder of mass m and radius r slides on a circular path of radiu

Question

A homogeneous cylinder of mass m and radius r slides on a circular path of radius R+r (i.e., the radius to the center of mass of the cylinder is R) without slipping on a large block that is fixed on the ground. Denote the angle of the location of the cylinder m with the vertical direction as . Assume the gravity constant g (a) What is the moment of inertia of this cylinder about its longitudinal axis (i.e., the direction out of the paper in the figure) passing through its center of mass? is the relationship between and ? AND the Lagrangian approach: (b) Denote the rotational angle of the cylinder as . Define that 0 when -0. What (c) Find the equations of motion of the cylinder in terms of using BoTH the Newtonian i. Newtonian approach: A. Denote the tangential force at the contact point between the cylinder and the block as F. Find the equation of motion for the translational motion in the tangential direction in terms of using F B. Find the equation of motion for the rotational motion in terms of using F C. Using part (b) and part (c)i.B, find the equation of motion for the rotational motion in terms of 1 D. Combine the above two equations of motion you found in part (c).i.A and part using (c).i.C by eliminating F. The solution should include only ; no or F A. Find the total kinetic energy in terms of and , and eliminate using part (b) This includes both translational kinetic energy and rotational kinetic energy B. Find the potential energy in terms of C. Find the Lagrangian of the system in terms of D. Find the equations of motion of the cylinder in terms of . Did it match what you found in part (c).i? (d) Consider the case where when is very small. Find the expression for the period of oscillation. How is this rigid body result different from the point mass case?

Explanation / Answer

given homogenous cylinder of mass m

radius r

radius of path = R + r

vertical angular location = theta

a. moment of inertia of cylinder about its longitudinal acis, Ic = 0.5mr^2

b. form the figure we can see

theta = arc/(R + r)

r*phi = arc

hence

theta = r*phi/(R + r)

c. i. Newtonian approach

let contact force between contact point and the cylinder be F

so, mgsin(theta) - F = ma

and

mv^2/(R + r) = mgcos(theta)

also, F*r = Ic*alpha

but alpha = d^2(phi)/dt^2 = (R + r)d^2(theta)/dt^2/r

also, a = r*d^2(phi)/dt^2 = r*alpha

hence

mgsin(theta) - 0.5m*a = ma

mgsin(theta) = 1.5ma

gsin(theta) = 1.5a

ii. Langragian approach

Total KE = 0.5mv^2 + 0.5Iw^2

v -> linear velocity

w -> v/r

Total KE = 0.5mv^2 + 0.25mv^2 = 0.75mv^2

Total PE = mgR(1 - cos(theta))

L = T -V = 0.75mv^2 - mgR(1 - cos(theta))

L = T -V = 0.75mr^2w^2 - mgR(1 - cos(theta))

d(dL/dw)/dt = dL/d(theta)

d(1.5*mr^2w - mgR(sin(theta)d(theta)/dw))/dt = 1.5mr^2w*dw/d(theta) - mgR( sin(theta))

1.5*mr^2dw/dt - mgR(sin(theta)d^2(theta)/dt*dw + cos(theta)*w*d(theta)/dw)) = 1.5mr^2w*dw/d(theta) - mgR( sin(theta))

now d(theta)/dw = w(dt/dw)

1.5*mr^2dw/dt - mgR(sin(theta)d^2(theta)/dt*dw + cos(theta)*w*d(theta)/dw)) = 1.5mr^2w*dw/d(theta) - mgR( sin(theta))

solving

gsin(theta) = 1.5a

d. for small osscillation case

g*theta = 1.5a

g*x/R = 1.5a

w = sqroot(g/1.5R)

for a point mass case,. w = sqroot(g/R)

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