rotation (urgent)

65) The hoop's centre of gravity is at a height of 3R above the rotation axis and using parallel axis thm, its M.I. about the axis is mR2/2 + m(3R)2 = 19mR2/2

Also the rod's M.I. about the axis is m(2R)2/3 = 4mR2/3, with its centre of gravity R above the axis.

So when then fall to upside down position:

GPE loss by the hoop = 2 x mg x 3R = 6mgR

GPE loss by the rod = 2 x mg x R = 2mgR

So assume that ω is the angular velocity at this moment, then:

(1/2)(19mR2/2 + 4mR2/3)ω2 = 8mgR

ω2 = 96g/(65R)

= 96.5

ω = 9.82 rad/s

67a) Chimney's M.I. about its base = mL2/3 where m and L are mass and length resp.

So at an inclination of θ to vertical:

GPE loss = mgL(1 - cos θ)

Assume that ω is the angular velocity at this moment, then:

mg(L/2)(1 - cos θ) = mL2ω2/6

3g(1 - cos θ) = Lω2

So sub θ = 35, we have the radial acc. Lω2 = 3 x 9.8 (1 - cos 35) = 5.32 m/s2

b) From 3g(1 - cos θ) = Lω2, taking diff. w.r.t. t:

3gω sin θ = 2ωL dω/dt

L dω/dt = (3g sin θ)/2 which is the tangential acc.

With θ = 35, linear acc. = 3 x 9.8 sin 35/2 = 8.43 m/s2

c) L dω/dt = g

(3g sin θ)/2 = g

sin θ = 2/3

θ = 41.8

2011-10-19 22:27:49 補充：
why d(3g(1 - cos θ))/dt = d(Lω2)/dt = 3gω sin θ = 2ωL dω/dt?
why there is a ω on the left side?

Since d(cos θ)/dt = - sin θ dθ/dt by using Chain rule, hence:

d(3g(1 - cos θ))/dt = 3g sin θ dθ/dt = 3gω sin θ as dθ/dt = ω (rate of change of angle = angular velocity)

65. Moment of inertia of the hoop about the axis = m(0.15)^2 + m(0.15+0.3)^2
= 0.225m
Moment of inertia of the rod about he axis =(1/3)m(0.3)^2 = 0.03m
Hence, total moment of inertia of the system = (0.225+0.03)m = 0.255m

Loss of potential energy of the system when it is upside down
= mg[2 x(0.15+0.3)] + mg[2x0.15] = 1.2mg = 12m
where g is the acceleration due to gravity, taken to be 10 m/s2

Therefore, use conservation of energy
12m = (1/2).(0.255)m.w^2
where wis the angular speed when the system is at the upside down position
solve for w gives w= 9.7 rad/s

67. (a) Loss of potential energy when the chimney is at 35 degrees with the vertical
= mg[L/2 - (L/2).cos(35)] = mg.(55/2).(1-cos(35)) J = 48.788m J
where g is the acceleration due to gravity, taken as 9.81 m/s2
Hence, by conservation of energy,
48.788m = (1/2).[m(55)^2/3].w^2
where [m(55)^2/3] is the moment of inertia of the chimney., and w is the angular acceleration at 35 degrees.
solve for w^2 gives w^2 = 0.09677 (rad/s)^2
Hence, radial acceleration = 55 x 0.09677 m/s2 = 5.32 m/s2

(b) Torque on the chimney = [mg.sin(35)].(55/2)
using torque = momentum of inertia x angular acceleration
Hence, [mg.sin(35)].(55/2) = [m(55)^2/3] x (angular acceleration)
i.e. (angular acceleration) = 0.1535 rad/s^2
Tangential acceleration of the top of the chimney
= 0.1535 x 55 m/s2 = 8.44 m/s2

(c) When the tangential acceleration is g, the angular acceleration = g/55
Hence, [mg.sin(theta)].(55/2) = [m(55)^2/3] x (g/55)
sin(theta) = 2/3
i.e. (theta) = 41.8 degrees