Straight Wire Slide

Classical Mechanics Level 3

A massive bead is positioned on a smooth straight wire whose ( x , y , z ) (x,y,z) end point coordinates are:

P 1 ⃗ = ( 1 , 2 , 3 ) P 2 ⃗ = ( − 15 , 5 , 7 ) \vec{P_1} = (1,2,3) \\ \vec{P_2} = (-15,5,7)

There is an ambient gravity of 10 m/s 2 10 \, \text{m/s}^2 in the − z -z direction. At time t = 0 t = 0 , the bead starts at rest at P 2 ⃗ \vec{P_2} and slides toward P 1 ⃗ \vec{P_1} .

At what time does the bead reach P 1 ⃗ \vec{P_1} ?

Note: All quantities are in SI units


The answer is 3.7483.

Steven Chase by Steven Chase , 37, USA

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

An energy analysis is easiest. The increase in kinetic energy is equal to the decrease in potential energy: 1 2 m v f 2 = − m g Δ z \tfrac12 mv_f^2 = -mg\Delta z . Thus v f = − 2 g Δ z . v_f = \sqrt{-2g\Delta z}. Since the acceleration is constant, the average velocity is the mean of initial and final velocity, i.e. ⟨ v ⟩ = 1 2 v f \langle v \rangle = \tfrac12v_f . Thus ⟨ v ⟩ = − 1 2 g Δ z . \langle v \rangle = \sqrt{-\tfrac12g\Delta z}.

The distance traveled by the bead is s = ( Δ x ) 2 + ( Δ y ) 2 + ( Δ z ) 2 ; s = \sqrt{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2}; finally, the time it takes to travel this distance is Δ t = s ⟨ v ⟩ = ( Δ x ) 2 + ( Δ y ) 2 + ( Δ z ) 2 − 1 2 g Δ z . \Delta t = \frac{s}{\langle v \rangle} = \sqrt{\frac{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2}{-\tfrac12g\Delta z}}. Substituting Δ x = 16 \Delta x = 16 , Δ y = − 3 \Delta y = -3 , Δ z = − 4 \Delta z = -4 , and g = 10 g = 10 , we obtain Δ t = 1 6 2 + ( − 3 ) 2 + ( − 4 ) 2 − 5 ⋅ ( − 4 ) = 281 20 = 14.05 = 3.7483 s . \Delta t = \sqrt{\frac{16^2 + (-3)^2 + (-4)^2}{-5\cdot(-4)}} = \sqrt{\frac{281}{20}} = \sqrt{14.05} = \boxed{3.7483}\ \mathrm{s}.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...