Frictionless block slides with speed 5 m/s along the road and enters on horizontal platform of a drawbridge?

Question

http://upload.wikimedia.org/wikipedia/commons/6/60/MovableBridge_draw.gif

The moment when the block eneters the platform, the platform comes into motion, raising at constant angualar speed 0.01rad/s.

At what elevation above the ground will the block come to rest wrt raising platform, and reverse direction of its motion?


Centripetal force is sufficiently small and can be ignored.

Answer 1

At an arbitrary position where the bridge is at an angle θ to the horizontal, the block is at r m from the starting position, travelling with radial velocity and acceleration r' and r" respectively.

Radial acceleration, ar = r" - rω^2
=~ r"

The only force acting in the radial direction is the component of the weight, mgsinθ

So r" = dv/dt = -g*sinθ
dv/dθ*dθ/dt = RHS
ω*dv/dθ = -g*sinθ

ω/g *dv = -sinθdθ
ω/g *(v-5)= cosθ - 1 ... (1)

v = 0 when θ = 5.79 deg = 0.101 radians

Now write v = dr/dt = dr/dθ*dθ/dt
= ω*dr/dθ

(v-5)= g/ω *(cosθ - 1)
dr/dθ = 5/ω + g/ω^2 *(cosθ - 1)
r = (5/ω)θ + (g/ω^2) *(sinθ - θ)

r(0.101) = 33.66 m

Vertical elevation = 33.66 *sin(0.101)
= 3.39 m

Note that the final potential energy is greater than the initial kinetic energy because work is required in order to elevate the block.

Answer 2

If we continue to assume frictionless surface,
mgh = (1/2)mv^2
h = (1/2)(5^2)/9.80665
h ≈ 1.3787 m ≈ 1.38 m.