Answer to Question #281864 in Mechanics | Relativity for Farhad Hassan

Question

A pinball machine launch ramp consisting of a spring and a 30∘ ramp of length 𝐿 as shown in Fig. 1.

(a) (3 marks) If the spring is compressed a distance 𝑥 from its equilibrium position and is then released at 𝑡 = 0, the pinball (a sphere of mass 𝑚 and radius 𝑟) reaches the top of the ramp at 𝑡 = 𝑇. Derive the expression for the spring constant 𝑘 in terms of 𝑚, 𝑔, 𝑥, and 𝐿. [Assume that the friction is sufficient, and the ball begins rolling without slipping immediately after launch.]

(b) (2 marks) What is the potential energy of the ball when it is at the midpoint of the ramp?

(c) (3 marks) Derive the expression of the speed of the ball immediately after being launched in terms of 𝑔 and 𝐿

Answer 1

Explanation & Calculation

• I assume you have the figure.

a)

• Assume also the pinball just makes it to the top of the ramp so that there is no speed or rolling of the ball at the top.
• Using the conservation of energy of the pinball from the point or release to the top of the ramp, a relationship can be written combining the elastic energy of the spring and the potential energy of the pinball.
• The stored elastic energy gives rise to the ball's potential energy.
• Therefore,

$\qquad\qquad \begin{aligned} \small \frac{1}{2}kx^2&=\small mgh\\ &=\small mg(L\sin30)\\ \small k&=\small \frac{mgL}{x^2} \end{aligned}$

b)

• The potential energy is,

$\qquad\qquad \begin{aligned} \small E_p&=\small mgh\\ &=\small mg(L\sin30)\\ &=\small \frac{mgL}{2} \end{aligned}$

c)

• The kinetic energy of the ball at a given point can be found as follows,

$\qquad\qquad \begin{aligned} \small KE&=\small \frac{1}{2}mv^2+\frac{1}{2}I\omega^2\\ &=\small \frac{1}{2}mv^2+\frac{1}{2}.\frac{2}{5}mr^2.\frac{v^2}{r^2}\\ &=\small \frac{7mv^2}{10} \end{aligned}$

• Then using the energy conservation from the compressed point to the just released point, the released speed can be found.

$\qquad\qquad \begin{aligned} \small \frac{1}{2}kx^2&=\small E_k\\ \small \frac{1}{2}\Big(\frac{mgL}{x^2}\Big)x^2&=\small \frac{7mu^2}{10}\\ \small u&=\small \sqrt{\frac{5gL}{7}} \end{aligned}$

• Note that if the ball does not just make it to the top, or it has some kinetic energy at the top, the solution is a whole different one.