Question

WORK AND ENERGY

The recently patented "Francis pile-driver" pounds piles into the ground by launching 15-kg bowling balls from a spring loaded gun, up around a frictionless loop-de-loop of radius r = 8 m, and down onto the pile. When the spring is compressed by delta s = 2 m the bowling balls have a speed of 20 m/s at the top of the loop, as shown. (You may assume g = 10m/s^2.) Find the work done by the frictionless track on a bowling ball as it goes from the gun to the top of the loop. Explain. Find the spring constant of the gun. Show your work. Find the acceleration of the bowling ball (magnitude and direction) at the top of the loop. Show your work. Find the net force (magnitude and direction) on the bowling ball at the top of the loop. Show your work. Find the normal force (magnitude and direction) exerted by the track on the bowling ball at the top of the loop. Show your work. Suppose the spring in the gun could be compressed by 4 m instead of 2 m. Would this cause the blowing balls to strike the piles with twice the speed, more than twice the speed, or less than twice the speed as before? Explain.

Answer 1

(a) Work done by frictionless track is zero since the normal force is always making an angle of 90 with respect to displacment and since cos90=0 we have work done=0

(b) Using conservation of energy

(c)

Radial accleraeration is given as =

$\\ a_r=\frac{v^2}{r} \\ a_r=\frac{20^2}{8} \\ a_r=50 \ m/s^2$ Downward

In this propblem component of tangential acceleration is very small and hence can be neglected

(d) Net Force at top

$\\ \sum F_{net}=N+mg=\frac{mv^2}{R} \Rightarrow \frac{15\times 20^2}{8} = 750 \ N$

(e)

$\\ \sum F_{net}=N+mg = \frac{mv^2}{R} \\ N=\frac{mv^2}{R}-mg \\ N=750-150 \\ N=600 N$

(f)

By conservation of energy

case 1

$\\ 0.5kx^2=0.5mv^2 \\ v=x\sqrt{\frac{k}{m}}$

case 2 x=2x

$\\ 0.5k(2x)^2=0.5mv_n^2 \\ 4kx^2=mv_n^2 \\ v_n=2x\sqrt{\frac{k}{m}} \\ v_n=2v$

So yeah on compressing the spring by twice more than before final impact velocity is doubled.